Edith Cowan University
Research Online
Theses: Doctorates and Masters
1995
On the relationship between a graph and the cycle
graph of its complement
Christian P. Lopez
Edith Cowan University
Recommended Citation
Lopez, C. P. (1995). On the relationship between a graph and the cycle graph of its complement. Retrieved from http://ro.ecu.edu.au/
theses/1184
This Thesis is posted at Research Online.
http://ro.ecu.edu.au/theses/1184
Theses
Edith Cowan University Copyright Warning You may print or download ONE copy of this document for the purpose of your own research or study. The University does not authorize you to copy, communicate or otherwise make available electronically to any other person any copyright material contained on this site. You are reminded of the following:  Copyright owners are entitled to take legal action against persons who infringe their copyright.  A reproduction of material that is protected by copyright may be a copyright infringement. Where the reproduction of such material is done without attribution of authorship, with false attribution of authorship or the authorship is treated in a derogatory manner, this may be a breach of the author’s moral rights contained in Part IX of the Copyright Act 1968 (Cth).  Courts have the power to impose a wide range of civil and criminal sanctions for infringement of copyright, infringement of moral rights and other offences under the Copyright Act 1968 (Cth). Higher penalties may apply, and higher damages may be awarded, for offences and infringements involving the conversion of material into digital or electronic form.
On the Relationship Between a Graph and the Cycle Graph of its
Complement
BY
Christian P. Lopez
B. Sc.
A Thesis Submitted in Partial Fulfilment of the
Requirements for the Award of
Master of S�ience ( Mathematics and Planning )
at the Faculty of Science, Technology
and Engineering, Edith Cowan University
November 10, 1995
USE OF THESIS
The Use of Thesis statement is not included in this version of the thesis.
2
Abstract
From an arbitrary graph G, another graph called the cycle graph of G and denoted by
C(G) can be derived. The cycle graph C(G) of G has as its vertices the chordless
cycles of G and two vertices in G(G) are adjacent if and orly if the corresponding
chordless cycles have at least one edge in common.
There has been a lot of work done on the cycle graph of a graph since its introduction
in 1984. This dissertation looks at the following open problem: Given a graph G, find
a relationship between G and the cycle graph C(G) of its complement. It is observed
that a vertex in C(G) corresponds to a chordless cycle in G and a chordless cycle
ck in G corresponds to a (k-3)-regular subgraph of order k of G, 3:, k:, IV(G)I,
Hence, the vertices of C(G) can be correlated to the structure of these (k-3)-regular
subgraphs of order kin G, and therefore a relationship between G and C(G) can be
established.
In this dissertation, the concepts of then-pointed star and the regular graph of a graph
are introduced. Ann-pointed star s11 corresponds to an (n-3)-regular graph of order
n, charncterised by some other properties emanating from it being a complement of
a chordless cycle. Chapter 3 explores some interesting features of this graph. On
the other hand, the regular graph R( G) of a graph G is an intersection graph whose
vertex set relates to the k-pointed star subgraph of G, 3::: k:, IV(G)I, and is later
shown to be isomorphic lo the cycle graph C( G) of the complement of G.
FinaJly, having found a relationship between G and C(G), the properties of the
resulting new map R : G
---+
C(G) are investigated. Several results are found
including the identification of the images under R of some special graphs such as
Nn, [{11 K(m1 n}, Sm Wn , Cn and P,l as well as graphs obtained by perfonning
,
some finite graph operations on then-pointed star and the complete graph. Also, the
behaviour r" these graphs under iterates of Rare examined and the minimum orders
for which Nn, Pn, <'-n, K(rn, n) and Sn are R-expanding, are determined.
3
Declaration
I certify that this thesis does not incorporate without acknowledgement any material
previously submitted for a degree or diploma in any institution of higher education;
and that to the best of my knowledge and belief it does not contain any material
previously published or written by another person except where due reference is
made in the text.
4
Dedication
To Dr. Evelyn L. Tan, who installed in me the love for Graph Theory, who unselfishly
shared with me her knowledge and experiences in conducting mathematical research,
and who gave me the inspiration to study hard to finish a masteral degree, this piece
of work is heartedly dedicated.
5
Acknowledgements
I would like to express my appreciation and thanks to Dr. Ute Mueller for supervising
me with my thesis; to Dr. Evelyn Tan for giving me the open problem for this thesis;
to my sister, Annabelle for the encouragement and for just being with me in Perth;
to my family back in the Philippines for being an inspiration to my thesis; to AusAid
for the scholarship that made my study here in Perth possible; to Glenn for the
support and for being a good friend; to John, Rod, Flor, Suresh and ail other friends
who made my two years of stay in Australia meaningful and unforgettable; and most
importantly, to HIM above for everything that I have achieved is not possible without
HIS guidance and help.
6
Contents
1 Introduction
8
Definition of Basic Concepts ................................................. 8
Background to the Cycle Graph of a Graph .................................. 10
Background and Approach to the Problem ................................... 12
Significance of the Study.................................................... 15
Some Results Significant to the Present Study ............................... 15
Some results on graph planarity .......................................... 16
Some results on cycle graphs ............................................. 16
2 On the Relationship Between a Graph and the Cycle Graph of its
Complement
19
Main Result. ................................................................ 19
Illustration ..................................................................26
3 On the n.-pointed star
30
4 On the Mapping R: G - C(G)
61
Properties of R . .... . . . . . . .. . .... . . .. . . . . . . ... . . . . . . . . . ...... . . . . . . ..... . . . . . 61
Identification of the Images Under R of Some Special Graphs ............... 67
Behaviour of Some Special Graphs Ur"'er Iterates of R ........... , ..........85
7
I
5 Summary, Discussion and Enumeration of Open Problems
95
6 List of References
99
8
Chapter 1
Introduction
Chapter I recalls the terms, concepts and results in graph theory significant to the
present study. This introductory chapter also clarifies the open problem considered
in this dissertation, discusses its origin and significance , and explains the approach
undertaken in the succeeding chapter to solve this problem.
Ll Definition of Basic Concepts
The following definitions are the basic graph-theoretical tenns that will be assumed
throughout the rest of this dissertation (Harary, 1969). Other concepts and notations
not included here will be introduced later on as they are needed.
A graph G
=
(V(G),E(G)) is an ordered pair of disjoint sets such that E(G) is a
set of 2-element subsets of V(G). The elements of V(G) are called the vertices of
G while the elements of E(G) are called the edges of G.
If e = nv E E(G), then u is said to be adjacent to 11 (symbolically, u,....., v), and the
vertices u and �, are incident to the edge e.
The cardinality IV(G)I of V(G), ie the number of vertices of G, is called the order
of G, and the cardinality IE(G)I of E(G), ie the number of edges of G is called the
size of G.
A graph G is finite, simple and undirected if IV(G)I is finite, G has no loops
or multiple edges and the edges do not indicate direction, respectively. All graphs
considered in this study are finite, simple and undirected.
9
A subgraph H of G is a graph having all its vertices and edges in G, ie V(H) �
V(G) and E(H) � E(G). A spanning subgraph Hof G is one which contains all
the vertices of G, ie V(H) = V(G).
For any subset S � V(G), the induced subgraph (S)0 is the subgraph of G with
vertex set Sand edge set {uv\ u,11 ES}.
The complement G of a graph G is the graph whose vertex set is the vertex set of
G and two vertices in G are adjacent if and only if they are non-adjacent in G.
Two graphsG 1 andG2 are isomorphic, writtenG 1 ,..,_,G2, if there exists a one-to-one
correspondence between their vertex sets which preserves adjacency.
A path Pn of order n is a sequence v 1v2 •••vn of n distinct vertices such that every
two consecutive vertices in the sequence are adjacent.
A cycle of length n is a sequence
1i1
v2... VnV1 of vertices where v1,1'2 1 • • • 1 Vn, n 2:'.: 3
are distinct and every two consecut!ve vertices in the sequence are adjacent. An edge
joining two non-consecutive vertices of a cycle is a chord of the cycle. A cycle of
length n without any chord is called a c hordless n-cycle and is denoted by Cn·
incident with u in G.
The degree of a vertex u in a graph G, denoted deg0 u, is the number of edges
An n�regular graph is a graph all of whose vertices have degree n.
A null graph Nn of order n is a 0-regular graph of n vertices, ie \V( Nn )I= n and
IE( Nn ) I= 0.
A complete graph Kn of order n is a graph consisting of n vertices and such that
�very pair of its vertices are adjacent.
A bipartite graph is a graph whose vertex set can be partitioned into two subsets
Vi
and "'2 such that each of its edges joins a vertex in
Vi with a vertex in '\.'2.
If
10
there is an edge joining every vertex in Vi to every vertex in V2 , then the graph is
called a complete bipartite graph. We denote this graph by K(m, n), if !Vil= m
and IV,I= n. A star Sn is the complete bipartite graph K(l,n).
Let S be a set and F={S1 , 82 , ... , Sn } a non-empty family of distinct non-empty
defined by V(Q(F)) = F, with S, and S; adjacent whenever i I j and S; n S; f 0.
subsets of S whose union is S. The intersection graph of F is denoted by O( F) and
12 Background to the Cycle Graph of a Graph
S. V. Gervacio (1984) introduced the concept of the cycle graph of a graph. He
defined the cycle graph C(G) of a graph G as the graph whose vertices are the
chordless cycles of G and two vertices in C(G) are adjacent if and only if the
corresponding chordless cycles have at least one edge in common.
For example, if G i s the graph
5
2
then its chordless cycles are a = 1231, b = 1241 and c = 1251. And since the edge
12 is common to all of these cycles, the cycle graph C( G) of G i s the triangle alica.
Given a graph G, the nth iterated cycle graph C"( G) of G is defined recursively
by C"( G) = C( C"- 1 ( G)), n e: 1 with C'( G) = G.
Returning to the above example, if we let x be the triangle abca in C( G), then G2( G)
is the graph consisting of just the vertex x and C 3 ( G) = 0.
Graphs can be classified according to their cycle graphs as either cycle-vanishing or
II
cycle-persistent. A graph G is called cycle-vanishing if there exists a non-negative
integer n such that C"(G)
=
0; otherwise, G is said to be cycle,persistent. A cycle­
persistent graph is further classified as cycle-periodic or cycle-expanding. A graph G
is cycle-periodic if there exist a positive integer p and a non-negative integer q such
that C'*'(G) = C'(G), where the smallest p is called the period of G. Otherwise,
G is cycle-expanding (Gervacio, 1984).
The graph G in the previous example is cycle-vanishing since C3 (G) = ©.
The complete graph of order 4, K4 , is cycle-periodic with period 1. This can be
shown as follows: Label the four vertices of K,i by 1, 2, 3 and 4. Then the chordless
cycles of K4 are a
=
1231, b
=
2342, c = 1431 and d = 1241, and every pair of
these cycles have a common edge. Therefore, in C(K,i), the vertices a, b, c, d are
pairwise adjacent so that C(K4 ) � K,.
The graph H below is cycle-expanding as it is neither cycle-vanishing nor cycle­
periodic.
If a
=
1231, b
=
2342, c
h = 2562, then C(H) is
=
1431, d
2
5
4
6
=
1241, e
=
a
b
e
t
c
d
g
h
And continuing in this manner, C2 {H} is
2512, f
=
5465, g
=
2642,
I
12
It is easy to see then that a graph G is cycle-expanding if and only if
J!..11i, IV( e ( G))I
n
= oo (Egawa, Kano and Tan, 1991 ).
Tan (1987) explored the inverse operation of associating to a graph G a graph H
with the property that its cycle graph is G, ie C(H) � G. In this case, G is called
a cycle graph and H its inverse cycle graph.
O>
ror example, K2 is a cycle gi:aph with H1 and H2 as two of its inverse cycle graphs.
,
H : rsr
H, :
In 1991, Egawa, Kano and Tan gave a charc1.cterisation of cycle-vanishing, cyc!e­
periodic and cycle-expanding graphs using the third iterated cycle graph of a given
graph. On the other hand, cycle graphs have not been fully characterised; a necessary
and sufficient condition for a graph to be a cycle graph has not yet been found. Le
and Prisner (1992) and Tan and Gervacio ( 1993) have characterised cycle graphs
inside certain small classes c.f graphs and have only produced results on necessary
conditions for a graph to be a cycle graph.
1.3 Background and Approach to the Problem
The investigation of the cycle graph of the complement of a graph was first proposed
by Tan (1993). One of the questions asked concerned that of finding a relationship
between the cycle graph C(G) of a graph G and the cycle graph C(G) of the
complement of G. However, there appears to be no relationship between C(G) and
C(G) since given two arbitrary distinct graphs, the isomorphism relation of their
cycle graphs and the isomorphism relation of the cycle graphs of their complements
are entirely independent of each other. For example, suppose that we have the
following graphs:
13
G, : lZ]
•
Then we get all the four possible situations:
C(G3 ) - C(G4) and C( c;3 ) - C( G4)
C( G1 ) "' C( G3 ) and C( G1 ) ,g C( G3)
,g C(G,) and C{G2) - C(G,)
C( G1 ) ,g C( G,) and C( Gi) ,g C( G2 )
C( G2 )
This means that for an arbitrary graph G, the structure of C( G) does not depend on
the structure of C(G).
On the other hand, if for any two graphs G 1 and G2 , G1 9:c: G2 , then it can easily be
deduced that G 1 "' G2 and that C( G 1 ) "' C( G2 ). Hence, the structure of C(G) is
related to the structure of G for an arbitrary graph G.
This dissertation investigates the relationship between a graph G and the cycle graph
C(G) of the complement of G.
It was observed that the vertex set and the edge set of G can be defined in tenns of
the vertex and edge sets of G, and the vertex and edge sets of C( G) can be defined
in terms of the structure of G, and hence of G. The approach undertaken to solve
the problem is based on the following analysis:
( 1 ) A vertex in C(G) is a chordless cycle, say
Ck = a,a2 . . . aka 1
in G, 3 :::::; k :::::;
IV( G)I , In G, Ck corresponds to a subgraph with the property that a1 ""' a2 , ak;
a2 ""'
a1 , a3j
a3
,t,, a2,a,1; ... ; ak-1 ,t,, ak-2 , a1.:; a1.: ""' a1, ak-I·
And since Ck is
chordless, each vertex in the subgraph is adjacent to the remaining k - 3 vertices
in G. Hence, if k is the length of the cycle, we have the fol1owing examples:
k
chordless cycles in G
0
3
ii
14
corresponding subgraph in G
•a
a4
al
o, ko,
aB
7
07
as
a2
- aa
6
•a 2
a3 •
02
0, 6 0 ,
5
l
oB
a,
07i
04
5
a4
al
as
a2
.h
a,
as
04
(2) Two vertices in C(G) are adjacent whenever the two corresponding chordless
cycles in G have an edge, say e
= uv,
i n common. In G, the two subgraphs
corresponding to the chordless cycles in G both contain u and v as non-adjacent
vertices.
From these observations, i t can be seen that the vertex and edge sets of C( G) can
be described i n tenns of the set of (k - 3)-regular subgraphs of order k of G,
3
:S
k
:S
!V(G)I such that there is a labelling in which consecutive vertices of
each subgraph are non-adjacent. Thus, the first step to solve the problem involves
characterising an arbitrary (k- 3)-regular subgraph of order k in G o f this type by
looking at its other properties originating from being the complement of a chordless
cycle in G. Then, using some graph theoretical results, it i s shown that there is
a one-to-one correspondence between the vertex set of C(G) and the set of these
(k - 3) -regular subgraphs of order k in G, 3
:S k :S
I V(G)I and that the adjacency
relation of the vertices of C(G) is related to these subgraphs in the manner described
in observation (2). An intersection graph whose vertex set relates to these ( k - 3)­
regular subgraphs of order k in G, i s defined and is shown to be isomorphic to C(G),
15
thereby establishing a relationship between G and C(G).
14 Significance of the Study
There have been no published results on the cycle graph of the complement of a
graph. It is hoped that this study will make some contributions to this subject and
hence may help in any further studies involving cycle graphs.
A closer look at the definition of the cycle graph C(G) of a graph G reveals that
C( G) is the edge -intersection graph of the set of all induced chordless cycles of G
(Tan, 1993). Although so far cycle graphs have not been applied in any other fields,
other intersection graphs are known to have been used as mathematical models in
representing some structural features: The circle graph has been used to model an
electric circuit diagram and to solve the problem of sorting a given pennutation on
a system of parallel stacks (Buckingham, 198 1 ). Unit disk graphs are used to model
broadcast networks and optimal facility location (Rosenkrantz, 1995). Intersection
graphs of planar objects have applications in image proces.sing, computer graphics,
circuit layout, and in certain type.s of scheduling problems (Fowler, Paterson, &
Tanimoto, 1981 ). Akiyama and Ruiz (1985) made an elementary survey of basic
types of intersection graphs such as triangulated graphs, circular-arc graphs and
vermutation graphs and discussed their uses in modelling problems.
LS Some Results Significant to the Present Study
This section presents a number of resu!ts on planarity and cycle graphs which are
necessary in the understanding of some of the concepts that will be discussed in this
dissertation and in the proofs of most of the lemmas, theorems and coro11aries that
will be given in Chapters 3 and 4.
The following results on planarity include a fundamental theorem and corollaries
16
on graph planarity which can be found in most basic graph theory books. On the
otner hand. the following results on cycle graphs include the most significant work
on cycle graphs such as Theorem 4 which gives a full characterisation of cycie­
vanishing, cycle-periodic and cycle-expanding graphs.
15.1
Some results on graph planarity
The following definitions and results are taken from Harary (1969):
A graph is planar if it can be drawn in such a way that no two of its edges intersect;
otherwise, it is called non-planar.
Two graphs are homeomorphic if and only if they both can be obtained from the
same graph by inserting a sequence of vertices into their edges.
Theorem 1 (Kuratowski's Theorem) A graph is planar if and only if it has no sub­
graph homeomorphic to !(5 or K(3, 3).
Corollary 2 Every planar graph G with IV(G)I 2 4 has at least four vertices of
degree not exceeding 5.
Corollary 3 /JG is any planar graph with IV(G) I 2 3, then IE(G)I S 3IV(G) j -6.
1.5.2
Some results on cycle graphs
A wheel W1, of order n + 1 is the graph consisting of a chordless n-cycle ei1 , n 2' 3
and a vertex u adjacent to all vertices of er,. An edge of Wn joining two vertices of
Cn is called an outer edge, and an edge joining u to a vertex of Cn is called a spoke.
A flip-flap wheel is a graph which consists of one wheel Wn and some 3-cycles
c3 such that Wn and c3 have exactly one edge in common, and every edge of Wn
belongs to at most one c3. We call c3 a flip or a flap according as the common edge
is a spoke or an outer edge of Wn · In particular, a wheel Wn is a flip-flap wheel
without flips or flaps (Egawa, Kano & Tan, 1969).
17
Theorem 4 (Egawa, Kano & Tan, I99I) let G be a graph. Then
3
( ! ) G is cycle-vanishing if and only ifC (G) is a null graph or an empty graph.
(2) G is cycle-periodic if and only if C3 (G) is not a null graph and each non­
trivial component of C3 (G) is a flip-flap wheel. In particular, the period of a
cycle-periodic graph is one or two; and
(3) G is cycle-expanding if and only if one of the nontrivial components of C3 (G)
is not a flip-flap wheel.
Theorem 5 below gives a sufficient condition for a graph to be cycle-expanding which
does not consider the third iterated cycle graph of the given graph, and instead locks
at specific types of subgraphs:
Theorem 5 (Egawa, Kano & Tan, 1991) If a graph G contains one of the graphs
Hn, H�, Jn and J�, n 2: 3, then G is cycle-expanding.
n
2
Hn
n
x
x
H'
4
n
2
x
y
J,
n
u
4
x
J'n
Theorem 6 (Egawa, Kano & Tan, 1991) Every non-planar graph is cycle-expanding.
The next theorem specifies a class cf graphs which are not cycle graphs.
18
Theorem 7 (Tan, 1987) let G be a graph for which G3 (G) fa 0 and G4 (G)
=
0.
Then G is not a cycle graph, ie there is no graph Lfor which C(L) '.:: G.
Let n. k, ,\ be positive integers such that n > k
> ,\.
The uniform intersection
graph G(n, k, A) is the graph whose vertices are all the subsets of cardinality k of
an arbitrary n-set, ie any set consisting of n elements; and two vertices are adjacent
in G (n, k, >.) whenever the corresponding k-subsets intersect in exactly ,\ elements
(Harary, 1969).
Theorem 8 (Tan, 1987) For n 2: 4, G(Kn) - G(n, 3, 2).
19
Chapter 2
On the Relationship Between a Graph
and the Cycle Graph of its
Complement
Chapter 2 gives the solution to the open problem considered in this dissertation, that
is a mathematical statement expressing the relationship between a graph and the cyc1e
graph of its complement; and applies this solution to a few examples. This chapter
also introduces the concepts of the n-pointed star and the regular graph of a graph
which will be studied further in the next two chapters.
2.1 Main Result
We start by characterising the (k - 3)-regular subgraph of order k of G, 3 ::; k
<
IV(G)I corresponding to the chordless k-cycle in G. In general, let us call it an
n-pointed star.
Definition 1 let n > 3. An n-pointed star
Sn
is a graph satisfying the following
properties:
( I) Sn is an (n - 3)-regular graph of order n.
(2) For n 2: 6,
Sn
does not contain K(1n, n - m) for any m, 3 :S m :S n - 3.
We will establish that for any n
2:
3, the n-pointed star is the comple:-nent of a
chordless n-cycle, and vice-versa. We will do so by first showing that the complement
of a chordless n·cycle is sn followed by a proof of the converse. But first, we define
a connected graph (Harary, 1 969).
20
A graph is connected if any two distinct vertices are joined by a path. A subgraph of
a graph G which is not contained in any other proper subgraph of G is called a com­
ponent of G, and 11:(G) denotes its number of components. Thus, G is disconnected
if s(G) > 1.
Proposition 9 The complement of a chordless n-cycle Cn is an n-pointed star sn,
n 2'. 3.
Proo£ Let Cn = a1a2 .. ,0-na1 be a chordless n-cycle. We v,ish to show that Cn � Sn,
Hence , we need to prove that Cn satisfies conditions ( 1) and (2) in the definition of
s•.
We first establish condition (I):
Since a1 a2 ...ana1 is a chordless n-cycle, each of the vertices a1, a2 1 . . . . a11 is adjacent
to exactly two vertices in en:
and for j = 2, 3, ... , n - l,
Thus, in c;;, each of these vertices is adjacent to exactly n - 3 vertices:
for j = 3, 4, ... n - 2,
Therefore, ({ a1 , a2 1 . . . 1 a 12} ) en fOims an induced (n - 3)-regular graph of order n, ie
c,. is an (n - 3)-regular graph of order n.
21
Next we show condition (2):
m < n - 3. Let {V,, V,} be the bipartition corresponding to K(m, n - m) where
Let n 2::: 6. Suppose, to the contrary, that Cn contains K(m, n - m) for some m, 3 �
6. Also, K(m, n - m) is a spanning subgraph of Cn since IVil + IV, I =
m + (n - m) = n = lcn l = lc,.I. These properties imply that ( Vi) ,. and ( V2) "' are
n
IVil
=
>
m and IVil
=
n - m. Note that V,, V, f 0, since 3 :S m :S n - 3 and
the two components of Cn so that Cn is disconnected. This i s a contradiction, since
by definition, an n-cycle is connected. Therefore, Cn does not contain K(m, n -m)
for any m, 3 :S m � n - 3 1 n 2: 6.
Therefore, Cn satisfies the definition of sn, ie the complement of Cn is Sn · •
To prove the converse of proposition 9, we need the next two lemmas.
Lemma 10 The complement Sn of an n-pointed star
Sn
is connected, n 2::: 3.
Proof: We have trivial cases for n = 3, 41 5 :
The 3 -pointed star
s3
consists of three pairwise non-adjacent vertices so that if s3 =
{a1 b, c} 1 then s3 is the triangle abc which is connected.
The 4-pointed star s4 is a pair of non-adjacent edges so that if s4 = { ab, cd}, then
s4 is the chordless 4-cycle adbca which is connected.
The 5-pointed star
s5
is a chordless 5 -cycle so that if
s5 =
acebda, then s5 is the
chordless 5-cycle abcdea which is connected.
2'.
x:{sn) > 1. Let H1 , . . . , Hp , p > 1 be the components of
Assume that n
regular,
sn
6. Suppose, to the contrary, that Sn is not connected. Then
Sn,
Since
Sn
is (n - 3)­
is 2-regular. Hence, it follows that for each i=l ...
, ,p, Hi is 2-regular
contains K(m, n - m), where Vi
and so IV( H;)I 2'. 3. Let IV( H1 )1
all other vertices of Sn, Thus,
sn
=
m. In
Sn ,
each vertex of H1 is adjacent to
=
V(H1) and
22
V, = V(sn)\V(H,) = U V(H,). Moreover, IV,I = I U V(H,)I = n - m ? 3
p
p
since IV(H,)I ? 3 for i;2, ... ,p. But n - m ? 3 is the same as n - 3 ? m so that
i=2
i=2
3 < m ::; n - 3; contradicting the definition of Sn· T herefore, sn is connected for
n > 6. •
A 2-regular graph of order n, n 2 3 may
n = 6, we may either have G 1 or G2 below:
0
be
connected or not. As an example, if
The following lemma tells us that if a 2-regular graph of order n, n 2 3 is connected,
then it is a chordless n-cycle.
Lemma 11 A 2-regular connected graph of order n, n 2 3 is the chordless n-cycle
c,..
Proo[ Let G be a 2-regular connected graph of order n, n ? 3 and V(G) ;
{a1 , a2 , . • . , an}· Then deg0 a;
=
2, for all i = 1,2, ...n. Without loss of generality,
Now, if a2 rv an , then a1 a2ana1 is the 3-cycle c3 and no other vertices in G can be
adjacent to c3 since deg0 a1
is disconnected or n
lemma holds.
=
deg 0 a2 = deg0 an
2. Thus, either n > 3 and G
3. By assumption, G is connected so that n
=
=
=
3, and the
If n > 3, then
a2
a vertex in
Thus, either n > 4 and G is dh:connected or n = 4. By assumption,
r;u an , Then there exists a vertex a3
# a1
an such that
since deg0 a1 == 2 and we. know that a1 "' a2 , an. If a3 rv an , then
a1 a2 a3ctna1 is the chordless 4-cycie c4 and nn other vertices in G can be adjacent to
Now. a3 r;u
a1
G is connected so that n == 4 and the lemma hoMs.
c4.
1
a2
rv
a3 •
23
Continue this process for n > 5 to generate the chordless cycle Cn· •
Proposition 12 The complement of an n-pointed star is a chord/ess n-cycle, n 2:: 3.
Proo£ By condition (]) of the definition of an n-pointed star sn and by Lemma
10,
is 2-regular of order n and is connected. Therefore, the proof immediately
follows from Lemma 1 1. •
Sn
Theorem 13 Let n 2:: 3. Chordless n-cycles and n-pointed stars are complementary
graphs.
Proo£ By combining Propositions 9 and 12, we have established the theorem. •
The definition of the complement of a graph implies that for any graph G, its com­
plement G always exists and is unique. The following corollary immediately follows
from this fact and the previous theorem.
Corollary 14 Let G be a graph and 3 '.'. k s; IV(G)I. There is a one-to-one corre­
spondence between the set of induced k-pointed star subgraphs of G and the set of
induced chordless k-cyc!es in G.
Proo£ For any induced k-pointed star subgraph of G, its complement uniquely
exists in G. By Theorem 13, it is an induced chordless k-cycle. Conversely, given
an induced chordless k-cycle in G, there uniquely exists an induced k-pointed star
subgraph in G. •
This corollary is fundamental to the proof of our main theorem- the theorem which
will define the relationship between a graph G and the cycle graph C(G) of the
complement of G. But fir&t, we introduce an intersection graph called the regular
graph of a graph.
24
Given G, let 3 S k S IV(G)I and define s<•> to be the union of an induced k­
pointed star subgraph sk and all pairs of non-adjacent vertices of sk.
If for fixed k,
G contains e. (ek 2: 2) induced k -pointed star subgraphs, we will write sj > , . . . , s):>
instead of
u u s,
IV(G)I ,,
k=3 i=1
Set F{G)
s<•>.
(k)
=
{sl'>, . . . , sJ:>1e. > o, 3 S k S IV{Gll} and S(G)
>
=
Definition 2 Given a graph G, ler F(G) and S(G) be as above. The intersection
graph of F(G) is called the regular graph R(G) of G.
identify the vertex set of R(G) with rhe set of induced star subgraphs of G, and two
Remark 1 From the definition of this intersection graph, it follows that we may
vertices in R(G) are adfacent
if the corresponding star subgraphs have a pair of
non-adjacent vertices in common.
Thus, we may restate the definition of our regular graph of a graph as follows:
Definition 3 Given a graph G, the regular graph R( G) of G is the graph whose
vertices are the k-pointed star subgraplzs of G, k=S 1 4 , ..., I V( G)I, and two vertices
in R( G) are adjacent if and only if the corresponding star subgraphs have at least
two non-adjacent vertices in common.
Example 1 Consider P5
:
2
P5 contains
{1,3 1 5} and
{12,45} as induced subgraphs. Hence, the
elements of V(R(P,)) are sl3> and sj l where
83
=
84 =
4
{ {1, 3, 5}, {1, 3}, {1, 5}, {3, 5}} and
{{12,45}, {1,4},{1,5}, {2, 4}, {2, 5}}.
Since sl3l n sl > = {1, 5} ,,;, 0, sl3l � s\4>. Thus, R(P,) is isomorphic to P,.
4
25
Example 2 Consider K3
:
b
c
As K3 does not contain any k-pointed star, 3 S k S IV(K,)I, we have V(R(K,)) = 0
so that R(K3 )
=
0.
Unless emphasis on the order of sk is required for clarity, we will refer to a k-pointed
star subgraph Bk simply as a star subgraph.
We now give our main theorem.
Theorem 15 l£t G be a graph. Then C(G) "' R(G).
Proo& By the definition of the cycle graph of a graph, C(G) is the graph whose
vertex set is the set of all chordless k-cycles, 3 S k S IV(G)I in G, and two vertices
in G( G) are adjacent if and only if the corresponding chordless cycles have at least
an edge in common.
By Corollary 14, there is a one-to-one correspondence between the set of all chordless
k-cycles in G and the set of all induced k-pointed star subgraphs
sk
of G, 3
:'.S
k ::-;
IV(G)I, Hence, it follows that the vertex set of C(G) corresponds to the set of all
induced k-pointed star subgraphs sk of G. Also, note that two chordless cycles in G
have an edge in common, say ab, whenever in the corresponding two star subgraphs
of G, a and b are non-adjacent. Thus, two vertices in C(G) are adjacent if and only if
in the corresponding two induced star subgraphs, there are at least two non-adjacent
vertices of G.
Now, we construct the set F( G) as above. Then there is a one-to-one correspondence
between the set of vertices of C( G) and F( G) and two vertices in C( G) are adjacent if
26
and only if the corresponding s)•>, sjm), (k, i)
cl (m, j) have an element in common:
si•> n sjml cl 0. T herefore, C(G) - R(G) by the definition of R(G). •
We give some more examples and explain the significance of our main result in the
succeeding section.
2.2 Illustration
In this section, we apply Theorem 15 to a few simple examples.
Example 3 Consider Ps again. We know that R(P5 ) '.::::::'. P2 . We now check this result
by getting P5 directly by definition, then its cycle graph C(P5) . P5 is
4
l
[1> 3
2
5
The induced chordless cycles of Ps are a = 14251 and b = 1351, and so C(P5 ) is
a
So we see that C(P5 )
�
R(P5 ) .
Example 4 Consider the
n below:
l
The elements o/ V(R(T6 )) are
sj'>
2
sl'>. sl3>, sj'>. sj'>, sj'i and si•> where
= {{1, 3, 5}, {1, 3}, {1, 5}, {3, 5})
27
sj31 sj'l st1 s�'l sj'I -
{{1,3, 6}, {l,3}, {1,6}, {3,6}}
{{1, 4,5}, {l, 4}, { 1, 5}, {4,5}}
{{1,4,6} , {1, 4}, {1,6}, {4,6}}
{{2, 4,6}, {2, 4}, {2, 6}, {4,6}}
{{34, 56}, {3, 5}, {3,6},{4,5}, {4,6}).
And R(T,) is
s
(3)
5
(3)
s'
Checking the result: T6 is
5
4
2
3
The induced chordless cycles of T6 are a
e = 2462, f = 35463, and so C(Tr,) is
6
= 1351, b = 1451, c = 1361, d = 1461,
e
d
Thus, C(T,) - R(T,).
28
Example 5 Let G be the graph below:
6
3
4
The elements of V(R(G))) are sj'I,
sj'I
sj'I
sj'I
Si51
-
sl'1, sj'I and s['1 where
{{6, 2, 4} , {6, 2} , {2 , 4} , {6,4}}
- {{56 , 23}, {5, 2}, {5, 3}, {6, 2} , {6,3}}
- {{16, 3 4}, { 1 , 3} , {1, 4} , {6, 3} , {6, 4}}
- {{123451}, {1 , 3} , {1, 4}, { 2, 4} , {2 , 5} , {3, 5} } .
Ariy two of these four vertices have a common element so that R( G) is isomorphic
to K4.
Checking the result: G is
4
2
6
5
3
utting a = 2462, b = 135241 , c = 25362 and d = 13641, we see that in C(G), a, b,
c and d are pairwise aqjacent so that C(G)is isomorphic to K,. Thus, C(G) � R(G).
Using our main result gives us an alternative way of finding the cycle graph of the
complement of a graph. Our previous examples did not exhibit much difference in
the effort exerted in applying the two methods considered. For certain complicated
graphs, however, our procedure will prove to be more handy and easier than applying
the actual definition, since the latter involves two steps: first, finding the complement
of the graph; second, finding the cycle graph of the new graph. Moreover, as seen
29
in the previous ex�mples, most of the simple graphs consist only of s3 , s4 and s5
which are three non-adjacent vertices, a pair of non-adjacent edges and a chordless
5-cycle, respectively, and are then not too difficult to determine in a graph. The
other k-pointed stars
sk
contain considerably more edges and possess some highly
structural features sur.:h as symmetrical star shape, and hence are found mostly in
large and complicated graphs.
30
Chapter 3
On the n-pointed star
In Chapter 2, we introduced the concept of the n-pointed star. The n-pointed star
appears to be an interesting graph as it has some special features. The following
properties are immediate consequences of its definition:
• It is a regular graph.
• IV(sn ) I = n and IE(sn ) I
= n(,� -3l.
• It does not contain a spanning complete bipartite graph with cardinality of bipartite
sets at least three as a subgraph.
Also, we know from Theorem 13 in Chapter 2,
• It is the complement of a chordless n-cycle.
This section explores some other properties of the n-pointed star which the author
believes to be significant within the context of the topic area. A major result of this
chapter includes a discussion of C(sn )· As we shall see later in Chapter 4, where
the characteristics of the new map R : G --+ C(G) are investigated, the images under
R of some special graphs are shown to b e related to C(sn ) = R(c,.).
The theorem below classifies the n-pointed star based on its cycle graph.
31
Theorem 16 The n-pointed star is cycle-vanishingfor n = 3, 41 5 and cycle-expanding
for n ?'. 6.
Proo£ The proof of this theorem requires a number of results on planarity of graphs
and tv:r"l results on cycle-expanding p:raphs given in Section 1.5.
Clearly, s3 and s4 do not contain cycles while C(s5) is a single vertex, so that they
are all cycle-vanishing.
For n = 6, we have
Thus, s6 contains the chordless cycles: A
a 1 a3a5a 1 , B
a2a4a5a2 i D = a1 asa2a4a1, E = a1 asa5a4a1 1 and so C (s5) is
=
= a3a5a2a5a3 1
C -
c
Now, note that C(so) � H3 of Theorem 5 so that C(s6 ) is cycle-expanding. Then
clearly, s6 is also cycle-expanding.
Assume that n � 7. By Theorem 6, every non-planar graph is cycle-expanding.
Thus, if we can show that for each n, n � 7, s n is non-planar, then we are done.
We first treat the cases where n = 7 �1nd n = 8.
32
o,
0
6
a,
o
a,
S7 : °7
a,
ss : 8
0
as
a2
a
7
as
a <'I
Observe that both s1 and s8 contain �
al
3
0
6
04
which is a graph homeomor-
phic to K(3, 3 ). Hence, by Theorem 1. s7 and s8 are n o n -planar and therefore, are
a5
a
6
o
7
cycle -expanding.
Now, for n > 91 we can also show that
Sn
contains a graph homeomorphic to
K( 3,3). However, we present an alternative proof which does not use the concept
of homeomorphism of graphs but instead makes use of Corollary 2: We note that
n > 9 implies thatn -3 > 5. So since sn is (n - 3)-regular of ordern by definition,
it follows that all vertices of Sn are of degree n - 3 > 5. Thus, Sn cannot have four
vertices of degree not exceeding 5. By Corollary 2, Sn is non-planar and is therefore
cycle-expanding for n 2:: 9. •
Note that so is the only n-pointed star which is planar and cycle-expanding. Also,
s6 is the onl)' n-pointed star which can be expressed as a product of two complete
graphs. To show this, we require the definition of the product of two graphs (Harary,
1969):
Two graphs are said to be disjoint if their corresponding vertex sets and their corre­
sponding edge sets are disjoint.
Let G1 and G 2 be two disjoint graphs. Their product, denoted by G1 x G2 , is the
graph with vertex set V(G1 x G2) ; V(G i ) x V(G2) and two vertices u = (u,,u,)
and v = ( 111, v2) in G1 xG2 are adjacent whenever u1
and U1
rv
V1,
=
111 and u2 ,..., 1J2 or u2 = v2
33
Now, we show that sa � K2 x Ks (triangular prism) as follows:
Ka :
6
b
c
Then V(K, x Ka ) = {(1 , a), (1, b), (1, c), (2, a), (2, b), (2, c)} and K2 x Ka is
(l.a)
(2.a)
(l,b
�
2. )
b
(l,c)
(2.c)
� (2.c)
(l. )
b
(l.a)
(2.a)
(2. )
b
(l.c)
f::f 85
For n > 6, the n-pointed star Sn is cycle-expanding by Theorem 16. Our goal is to
give some characteristics of the cycle graph C(sn) of sn, n > 6 by examining the
structure of the chordless cycles of sn and their relationship with each other.
Firstly, we look at the length of the chordless cycles in sn , n > 6. Theorem 18 tells
us that any induced chordless cycle in Sn is of length either 3 or 4.
Lemma 17 If Sn colltains c5 as an induced subgraph, then n = 5 and Sn = c5 .
Proof. By the definition of the complement of a graph and by assumption, it follows
that sn contains cs as an induced subgraph, n
2. 3. By Theorem
13,
Sn
= Cn and
c5 = s5 so we have Cn contains s5 = c5 . And so since Cn is chordless, it follows that
Cn = c5, ie n = 5 so that Sn = c5 . •
Theorem 18 Let ck be an induced chordless k-cycle contained in sm 3
:S
k
<
n, n 2:: 6. Then k = 3 or 4.
Proo£ Let u be an arbitrary vertex in ck . Then degck u = 2 and u is non-adjacent
to exactly k - 3 vertices in ck, Since Ck � Sn , <legs" u 2:: 2 and u is non-adjacent to
at least k - 3 vertices in Sn , This means that in sn , u is adjacent to at least k - 3
34
vertices, ie <legs; u > k - 3. By Theorem 13,
= £;i so that every vertex in
is of degree 2 which implies that degsn u = 2. Hence, we have 2 ?: k - 3 so that
k S 5. But k
Sn
Sn
# 5, otherwise by Lemma 17, n = 5 which is a contradiction to the
assumption. Therefore, k = 3 or 4. •
We will now examine the degree of the vertices in C(sn ), n
?: 6. As we know, a
vertex C* in C (sn) corresponds to an induced chordless cycle in Sn. Thus, in relation
to Theorem 18, there are two cases to be considered: ( 1 ) C* corresponds to a 3-
cycle cj in sn , and (2} C* corresponds to an induced chordless 4-cycle
c4
in Sn·
Moreover, by the definition of the cycle graph of a graph, the degree of C* i n G(sn}
corresponds to the number of induced chordless k-cycles in sn, 3 :::; k S n, which
have an edge in common with c; , i = 3,4. And again by Theorem 18, k = 3 or 4
so that degC(s., ) C* is the sum of the number of 3-cycles in
which have an edge
in common with c;, and the number of induced chordless 4 -cyc\es in
an edge in common with
C:, where either i = 3 or i = 4.
Sn
We treat first the case where C* corresponds to a 3-cycle c3 in
Sn
which have
Sn·
Recall that the distance da (u, v) between two vertices u and v in a graph G is
the length of a shortest path joining them (Harary, 1 969). The following lemma
enumerates the number of induced 3-pointed stars in a chordless cycle containing
two specific vertices in terms of the distance between them.
Lemma 19 Given two distinct non-adjacent vertices u and v in a chordless n-cycle
en, n ?: 6, the number a (u, v) of 3-pointed stars containing u and v is
a{u, v) = {
n - 5 , ifd,,(u, v) = 2
n - 6, if d,)u, v) > 2
Proo£ Since u and v are distinct, den (u, 11)
in Cn, d,Ju,v) # 1 so that d0n (u,v) > 2.
f. 0, and since u and v are non-adjacent
Suppose that dc,. (u, v) = 2. Let Cn = 1 2 ...i-2 i-1 i=u i+l i+2=1J i+3 i+4...nl, n � 6.
35
2
�� ',
1-2
1+3 �11-1
l=U
1+2=V
I+ l
1+4 ·
Then the 3-pointed stars containing u and v are of the form {j, u, v} , j = 1, 2, ... 1 n,
j =j:. i-1, i, i+1, i+2, i+3 and so if a(u, v) is the number of s3 in Cn containing u and
v, then a(u, v) = n - 5.
Suppose that d,. (u,v) = k > 2. Let Cn = 1 2...i-2 i-1 i=u i+l i+2...i+k-2 i+k-1
i+k=v i+k+l i+k+2 ...n, 3 S k S n-i, n 2: 6.
n ,..._.-• . .
1
1-2
i-1
l
l+k+2
l+k+l
2
l
i+k=V
l+k-1
i=U
i+ l
,'
l+k-2
:
1+2
Then the 3-pointed stars containing u and v are of the form {j, u, v}, j = 1,2,... ,n,
j ,f i-1, i, i+l, i+k-1, i+k, i+k+l and so o:(u, v) = n - 6. •
We may thus compute the number of 3-pointed stars which have two vertices in
common with a given fixed 3-pointed star in a chordless cycle:
Lemma 20 Given an arbitrary 3 -pointed star s3 = { a, b, c} in a chordless n-cycle,
n � 6, the number a(s3) of 3 -pointed stars s3 =/:- sj in Cn which have exactly two
vertices in common with sj is
et(s3) =
36
0,
3n - 19,
3n - 20,
3n - 21,
if dc,, (a,b) = 2, de,, (b,c) = 2 and dc,, (a,c) = 2, n = 6
if dc,, (a, b) = 2, de,, (b, c) = 2 and d,.(a, c) > 2, n ?>:: 7
if d,,, (a,b) = 2, de,, (b, c) > 2 and dc,, (a, c) > 2, n > 8
if de,, (a, b) > 2, de,, (b, c) > 2 and d"' (a, c) > 2, n ?>:: g
Prout: Observe that in a cycle with any three non-adjacent vertices a, b, c, inter­
changing the labelling of these vertices such as a --+ b, b - c, c --+ a, results in a
cycle containing a, b, c which is isomorphic to the original cycle under the isomor­
phism in which a --+ b, b --+ c, c --+ a. Hence, for the proof of Lemma 20, we only
look at the following which give the non-isomorphic cases:
Case I.
de,, (a, b) = 2,
de,, (b, c) = 2 and de,, (a, c) = 2
Case 2. dc,(a,b) = 2, d"' (b,c) = 2 and dc,, (a,c)
Case 3. d,,, (a, b) = 2, d"' (b, c)
Case 4. d,,, (a,b)
>2
> 2 and de, (a, c) > 2
> 2, de, (b, c) > 2 and dc, (a, c) > 2
Now, by Lemma 19, the number of 3-pointed stars in en which have exactly two
vertices in common with sj = { a, b1 c} is
<>(s;) - [a(a, b) - l] + [a(b, c) - l] + [<>(a, c) - l]
- a(a, b) + a(b,c) + <>(a,c) - 3
(•)
where subtracting the value 1 from every summand corresponds to the fact that sj
is counted in each of these terms and thus needs to be removed.
Case I. d,,, (a, b) = 2, de,, (b, c) = 2 and de,, (a, c) = 2
37
Then n = 6 and by (•) and Lemma 19, we have a(sl) = 3(6 - 5) - 3 = 0, ie there
are no 3-pointed stars in
C(3
with exactly two vertices in common with sj. This can
also be observed in the figure above: the only s3 containing any two of the vertices
in {a1 b, c} is sj.
Case 2. d,,. (a,b) = 2, d.,,, (b,c) = 2 and d,,Ja,c) > 2
a
.
b
c
Then n > 7 and by (•) and Lemma 19, we have a(s3 ) = (n - 5) + (n - 5) + (n 6) - 3 = 3n - 19.
Case 3. dc,, (a,b) = 2, d.,,, (b,c) > 2 and d,. (a, c) > 2
a
O
b
c
Then n 2: 8 and by (•) and Lemma 19, we have a(s3 ) = (n - 5) + (n - 6) + (n 6) - 3 = 3n - 20.
Case 4. d,,,. (a, b) > 2, d,,. (b, c) > 2 and d,,,. (a, c) > 2
Then n 2: 9 and by (•) and Lemma 19, we have a(s3 ) = (n - 6) + (n - 6) + (n -
38
6) -3 = 3n - 21. •
The next lemma gives the number of 3 -cycles in Sn which have an edge i.n common
with c;, where c; is a 3-cycle in Sn corresponding to a vertex
Lemma 21 Let c; = abca be an arbitrary 3-cycle in
Then the number ci( cj) of 3-cycles c3 -=/= cj in
with c; is
O,
o/(c;) =
Sn
Sn,
c; in C( sn), n > 6.
n � 61 and let Sn = Cn·
which have an edge in common
ifdc.. ( a, b) = 2, de.. ( b, c) = 2 and de.. ( a, c) = 2, n = 6
3n - 19, ifde.. ( a, b) = 2, de., ( b, c) = 2 and de. ( a,c) > 2, n � 7
3n - 20, if dc., ( a, b) = 2, de.. ( b, c) > 2 and de. ( a, c) > 2, n � 8
3n - 21, if d�. ( a,b) > 2, de., ( b, c) > 2 and de.,( a, c) > 2, n � 9
Proo£
Sn
= Cn by Theorem 13. Theorem 13 also implies that cj = abca in
Sn corresponds to a 3-pointed star sj = {a, b, c} in Cn· Furthennore, for every 3 cycle c3 in
Sn
which has an edge, say ab in common with cj, there is an induced
3 -pointed star s3 in Cn which contains a and b as non-adjacent vertices so that
V( s3 ) n V( s3 ) = {a, b), and vice -versa. Hence, the result i mmediately follows from
Lemma 20. •
We next compute the number of induced 4 -pointed stars in a chordless cycle con­
taining two distinct fixed vertices.
39
Lemma 22 Given two distinct non-adjacent vertices u and v in a chordless n-cycle
Cn, n 2: 61 the number {J(u, v) of induced 4-pointed stars s4 in Cn containing u and
11
is
1, ifd.,,, (u, v) = 2, n 2 6
f3(u,,i) =
2, ifd.,,, ( u, v) = 3, n = 6
3, ifd.,,, (u, v) = 2, n 2 7
4 , if dcJu,v) 2 4, n 2 8
Proot: Since u and v are distinct and non-adjacent in Cn, dc,, (u,v)
2: 2. Let {J( u,1i)
be the number of induced 4-pointed star s,1 in Cn containing u and v.
Suppose that dc,, (u1 v)
=
2. Also, by assumption, n 2: 6 so we have the following
situation:
Then the only induced 4-pointed star in Cn containing u and v is {u3,v2} so that
f3(u, ,i) = 1.
Suppose that d0,(u, 1J) = 3.
If n = 6, ea is
43
1
0
2
v
and the induced 4-pointed stars s4 in C(i containing u and v are {u3, v2} and {ul, v4}
40
so that {3(u, v) = 2.
If n 2: 7,
c. is
- ointed stars in Cn containing u and v are {u4, v2}, {u4, v3} and
and the induced 4 p
{ul,v3) so that {3(u,v) = 3.
Suppose that de,. (u, v) � 4, then Cn is
�
4 3/
\1 2
�5
Note that this case only occurs when n � 8. And the induced 4-pointed stars in Cn
containing u and ,, are {u3,v5), {u3, ve), {ul,v5) and {ul, 1!6} so that {3(u,v) =
4. •
We may use Lemma 22 to count the number of 4-pointed stars in a chordless cycle
which have two non-adjacent vertices in common with a fixed 3-pointed star.
41
Lemma 23 Given an arbitrary 3-pointed star sj = {a, b,c} in a chordless n-cycle
vertices in common with s; is
Cn, n � 61 the number /3(s3) of 4-pointed stars 84 in Cn which have two non-adjacent
3,
5,
6,
,B(s;) =
7,
ifd,.(a, b) = 2, dc,, (b,c) = 2 and dc,, (a,c) = 2, n = 6
if d,. (a, b) = 2, d,. (b, c) = 2 and d,. (a, c) = 3, n = 7
if de,, (a, b) = 2, d,. (b, c) = 2 and d,. (a, c) 2: 4, n 2: 8
ifd,,,(a,b) = 2, d,. (b, c) = 3 and d,,,(a,c) = 3, n = 8
if d,. (a, b) = 2, d,. (b, c) = 3 and d,.(a, c) 2: 4, n 2: 9
9, if [d,. (a, b) = 2, de.. (b, c) 2: 4 and d,. (a, c) 2: 4, n 2: 10)
8,
or [dc,, (a,b) = 3, d,. (b, c) = 3 and d,,,, (a, c) = 3, n = 9)
10, 1f de. (a, b) = 3,
de,, (b, c) = 3 and d,,, (a, c) 2: 4,
n 2: 10
11, if d,. (a, b) = 3, d,. (b, c) 2: 4 and rl,,, ( a, c) 2: 4, n 2: 11
12, if d,. ( a, b) 2: 4, d,,, (b, c) 2: 4 and d,,, (a, c) 2: 4, n 2: 12
Proo[ Let fJ( sj) be the number of 4-pointed stars in en which have two non-adjacent
vertices in common with s; = { a, b, c}, then /3(s3 ) = (!( a, b) + ,B(b, c) + /3(a, c) by
Lemma 22. Also, with reference to Lemma 22, the non -isomorphic cases that need
to be considered are the following:
Case ], d,. (a,b) = 2, d, (b,c) = 2 and rl,,(a,c) = 2
0
By Lemma 22, f3(a, b) = 1, ,B(b,c) = 1 and ,B(a,c) = 1 so that /3(s3 ) = 3. To
determine the order of Cn, we impose the conditions in Case I to
a
en
to get
cob
and thus, n = 6.
Conversely, if n = 6, any 3 -pointed star {u1 v 1 w } in Cn has the property that
dcJu,v) = den (v,w) = dcJu, w) = 2. Hence, in the succeeding cases, n ?: 7.
42
Case 2. d," (a,b) = 2, do,, (b,c) = 2 and ds, (a, c) = 3
By Lemma 22, (J(a, b) = l, (J(b, c) = 1 and (J(a, c) = 3 so that (J(sl) = 5. We get
c,, as
and thus, n = 7.
Case 3. d,Ja, b) = 2, d," (b, c) = 2 and d0,{a,c) 2: 4
By Lemma 22, (J(a, b) = l, (J(b, c) = 1 and (J(a, c) = 4 so that (J(s3)
c,, as
=
6. We get
a
b
c
and thus, n > 8.
Case 4. d," (a,b) = 2, do,, (b,c) = 3 and d," (a,c) = 3
By Lemma 22, (J(a, b) = l, (J(b, c) = 3 and (J(a, c) = 3 so that (J{sl) = 7. We get
c,, as
and thus, n = 8.
43
Case 5. dc,, (a,b) = 2, den (b,c) = 3 and de,, (a,c) 2 4
By Lemm a 22, (J(a,b) = l, (3(b,c) = 3 and (J(a,c) = 4 so that (J(s;) = 8. We get
c,, as
a
b
c
and thus, n 2 9.
Case 6. d,. (a,b) = 2, d,,. (b,c) 2'. 4 and d,,. (a,c) 2 4
By Lemma 22, f3(a,b) = 1 , /3(b,c) = 4 and (J(a,c) = 4 so that f3(s3 ) = 9. We get
c,, as
� - ··
and thus, n 2 10.
Case 7. d"" (a, b) = 3, d,,. (b, c) = 3 and den (a, c) = 3
By Lemma 22, (3(a,b) = 3, (3(b, c) = 3 and (J(a,c) = 3 so that (3(s3 ) = 9. We get
c,, as
44
and thus, n = 9.
Case 8. dc,, (a, b)
= 3, de. (b,c) = 3 and dc,, (a,c) 2: 4. By Lemma 22, (J(a,b) = 3,
{3(b, c) = 3 and f3(a,c) = 4 so that f3(s3) = 10. We get Cn as
a
b
c
and thus, n
2: 10.
Case 9. de.(a, b) = 3, de. (b, c) 2: 4 and de,, (a, c) > 4
By Lemma 22, f3(a, b) = 3, f3(b, c) = 4 and (J(a, c)
Cn as
= 4 so that f3(s3) = 11. We get
a
b
�. · ·
and thus, n 2: 11.
Case JO. d�. (a,b)
2: 4,
de. (b,c) 2: 4 and dc,, (a,c)
By Lemma 22, f3(a, b) = 4, f3(b, c) = 4 and f3(a, c)
Cn as
2: 4
=
4 so that (3(s3 ) = 12. We get
45
and thus, n 2 12. •
The succeeding lemma gives the number of induced chordless 4-cycles in
which
have an edge in common with cj, where cj is a chordless 3-cycle in sn corresponding
to a vertex Cj in C(s11 ), n � 6.
Sn
n � 6 1 and let Sn
Then the number f3'(cj) of induced chordless 4-cycles c4 in Sn which have an edge
Lemma 24 Let c; = abca be an arbitrary 3-cycle in
in common with cj is
5,
3,
6,
/l'(c; )
=
7,
8,
9,
10,
11,
Proot: sn
12,
if de,, (a, b) = 2,
if de,Ja,b) = 2,
Sn,
=
Cn·
d,,, (b, c) = 2 and de,, (a, c) = 2, n = 6
do. (b, c) = 2 and de.. (a, c) = 3, n = 7
ifde,,(a,b) = 2, d,,, (b, c) = 2 and do. (a,c) z 4, n 2'. 8
if den (a,b) = 2, d,,, (b, c) = 3 and den (a,c) = 3, n = 8
if den (a, b) = 2, d,.. (b, c) = 3 and d�. (a, c) 2 4, n 2 9
if [den (a,b) = 2, d,,, (b, c) 2 4 and d,,, (a, c) 2 4, n 2 10]
or [de,, (a, b) = 3, do. (b, c) = 3 and deJa, c) = 3, n = 9]
if de (a, b) = 3, d,,, (b, c) = 3 and de,, (a, c) 2 4, n 2 10
ifrJ,,,.(a,b) = 3, d,,,, (b, c)
,,
if den (a, b) 2 4,
z 4 and d,,,,(a, c) z 4, n 2 ll
d,,, (b, c) 2 4 and d"' (a, c) 2 4, n 2 12
Cn by Theorem 13. Theorem 13 also implies that cj
abca in
corresponds to {a, b, c} in Cn· Furthermore, for every induced chordless 4-cycle in Sn
=
=
Sn
which has an edge, say ab in common with c,3, there is an induced 4-pointed star in en
which contains a :.md b as non-adjac0nt vertices so that V(sj) n V(s4) = {a, b}, and
46
vice-versa. Hence, the result immediately follows from Lemma 23. •
Now, we combine the results of Lemmas 21 and 24 to get degG(sn)
a vertex in C(sn) corresponding to a 3-cycle c; in SnTheorem 25 let
c;
1
where 03 is
c; be a vertex in C(sn) corresponding to a 3-cycle c; = abca in
Sn, n � 6 . If Sn = Cn, then degO(sn) Gj has the Value
if dc,. (a, b} = 2, d," {b, c} = 2 and dc,. {a,c} = 2, n = 6
3,
if d0,(a, b} = 2 , de,, (b. c} = 2 and d,o,, {a,c) = 3, n = 7
7,
if dc,. (a,b} = 2 , d," (b,c) = 3 and dc,. (a,c) = 3, n = 8
11 ,
if dc. (a,b) = 3, d," (b,c) = 3 and dc,. (a,c) = 3, n = 9
15,
if dc,. (a,b} = 2, d," (b,c) = 2 and d,Ja,c) 2 4, n > 8
3n - 13,
3n - 12, if dc. (a,b) = 2, d,,, (b,c) = 3 and d,,, (a,c} 2 4, n > 9
3n - 11, if [de,. (a, b} = 2, d," (b, c} 2 4 and de,, (a, c} 2 4, n 2 10]
or [dc,,(a, b} = 3, d,,, (b, c) = 3 and d"' (a, c} 2 4, n > 10]
3n - 10, if dc,. (a, b} = 3, d," (b, c} 2 4 and dc,. (a, c} 2 4, n > 11
if d," (a, b} 2 4, d,," (b, c} 2 4 and d,,,, (a, c) 2 4, n 2 12
3n - 9,
Proo£ By Lemma 21 and Lemma 24, degc(s,,) G';; = the number of c3
i
c3 in
sn which have an edge in common with cj + the number of induced chordless
4-cycles in Sn which have an edge in common with cj = a'(cj) + {3'(cj). Combining
the values obtained in these two lemmas, we get o:'(cj) + {3'(cj) equal to
3,
3n - 14,
3n - 13,
3n - 12,
3n - 11,
3n - 10,
3n - 9,
if dc,. (a, b) = 2, de,, (b,c} = 2 and dc,, (a, c) = 2, n = 6
if dc,. {a,b} = 2, de,, (b,c} = 2 and d0, (a,c) = 3, n = 7
if [d0,(a, b}
= 2, d,. (b, c} = 2 and d,. (a, c} 2 4, n 2 8]
or [dc,. (a,b} = 2, d,. (b,c) = 3 and den {a,c} = 3, n = S]
if [dc,,(a, b} = 2, d,. (b, c} = 3 and den (a, c) 2 4, n 2 9]
or [dc,. (a,b} = 3, den {b,c} = 3 andd,.{a, c} = 3, n = 9]
if [den (a, b) = 2 , d," (b, c} 2 4 and de,. (a, c) 2 4, n 2 10]
or [dc,. {a,b) = 3, dc,, {b, c} = 3 and d,. {a, c) 2 4, n :0: 10]
if de,. (a, b) = 3, den (b, c} 2 4 and de,. (a, c} 2 4, n 2 11
if den (a, b) 2 4, d,. (b, c} 2 4 and de,. (a, c) 2 4, n 2 12
We may further simplify the value of oi (cj} + /3'(c3) in some of these cases: If
47
den (a,b) = 2, den (b,c) = 2 and d,.(a,c) = 3, n = 7 so that 3n - 14 = 7. For
den (a,b) = 2, den (b,c) = 3 and d'" (a,c) = 3, n = 8 so that 3n - 13 = 11; and
finally, for den (a,b) = 3, den (b,c) = 3 and d,.(a,c) = 3, n = 9 so that 3n- 12 = 15.
Then the result follows. •
Now, we treat the case where G* corresponds to an induced chordless 4-cycle c4 in
Sn.
As in the first case, we present some preliminary lemmas.
Lemma 26 Given an arbitrary induced 4-pointed star s;;_ = {ab, cd} in Cn, n 2 6,
the number 1(s4) o/3 p
- ointed stars in Cii which have exactly two vertices in common
with s4, is
2,
5
7(s4) =
4n - 23,
4n - 24,
Proo£ In
if d,. (a,c) = 2,
if d,0 (a,c) = 2,
if d,,, (a,c) = 2,
if d,. (a,c) > 2,
d,. (a,d) = 3, d,0 (b, c) = 3, den (b, d) = 2,
n=6
d,. (a,d) = 3, d,, (b, c) = 3, d,.(b, d) = 3,
n=7
de,, (a,d) = 3, d," (b, c) = 3, d,,(b, d) = 4,
n 2: 8
d,,, (a,d) > 2, d,,, (b, c) > 2, d.,,, (b, d) > 2,
n 2: 8
s4 = {ab1 cd}, we know that a ,....., b; c ,...., d; a r,t, c, d; b r,t, c, d so that the
3-pointed stars sa which have exactly two vertices in common with s,! are of the form
only pairs of non-adjacent vertices are {a, c), {a, d), {b,c) and {b,d). Hence, the
{a1 c1 x},x
'"JU
a,c; {a, d, x},x
r,t,
a,d; {b1 c,x} 1 x
,,v
b1 c; and {b, d, x},x
r,t,
b, d. If
follows from Lemma 19 that the number of such s3 depends on whether the distance
between the two non - adjacent vertices in each pair is equal to or greater than 2. We
therefore obtain the following cases up to isomorphism:
Case I. d,.(a,c) = 2, d,. (a,d) = 2, d,,, (b,c) = 2 and dc,, (b,d) = 2
Case 2. den (a, c) = 2 , d.,,, (a, d) = 2, d," (b, c) = 2 and d"' (b, d)
>2
48
Case 3. de,, (a,c) = 2, d"' ( a, d} = 2 , de,. (b, c} > 2 and dc,. (b, d) > 2
Case 4. dc,, (a, c) = 2, d"' (b,d} = 2 , de,. (b, c) > 2 and d,.(a, d) > 2
Case 5. dc,, (a,c) = 2, d"' (a,d) > 2 , d,. (b, c) > 2 and d,,,, (b, d) > 2
Case 6. d,.(a,c) > 2, d,. (a,d} > 2 , d,. (b,c} > 2 and dc,. (b,d) > 2
Let 'Y(s.;i) be the number of s3 in Cn which have exactly two vertices in common
with s4, then 1(s4) = a(a, c) + a(a, d) + a(b, c) + a (b, d). We determine its value
for the different cases.
First, we consider the case when den (a,c) = 2 and dc,,(a,d) = 2. We have the
following situations:
b
a
(i) (]
b
c
d
(ii) <S
d
c
a
a
b
c
d
(iii) LJ
(i) implies that b ,...., d; (ii) implies that either Cn is not a cycle or Cn has a chord;
and (iii) implies that b ,...., c. These are all contradictions to the assumption so that
d,,,, (a, c) and d,. (a, d) cannot be both equal to 2.
Now, consider the case where d,. (a,c} = 2 and dc,, (b,d} = 2. Then we have the
following possibilities:
a b
a b
(i) G or S
c
d
c
d
a
b
c
d
(ii) o
Z
a
b
a
b
c
d
c
d
(iii) Z>or
(i) implies that a "' d while (iii) implies that b rv c which are both contradictions to
the assumption. On the other hand, (ii) satisfies all the assumptions. Hence, we now
know that whenever de,. (a, c) = 2 and d"' (b, d) = 2, then n = 6, d,. (a, d) = 3 and
d0,(b,c) = 3.
49
By Lemma 19, a( a, c} = 6 - 5 = I, a( a,d) = 6 - 6 = 0, a( b, c) = 6 - 6 = 0 and
a(b,d} = 6 - 5 = I so that 7(s4 ) = I + o + o + I = 2.
From the foregoing discussion, we deduce that Cases ], 2 and 3 are not possible and
that Case 4 reduces to dc,. ( a,c) = 2, de,. ( b, d} = 3, d,,. ( b, c)
=
3 and dc,. ( a, d} = 2,
n = 6.
We examine the remaining cases.
Case 5. d,,. (a, c) = 2, d,,. (a, d} > 2, d,,. ( b, c} > 2 and d,,. ( b, d}
>2
Then the position of s4 in Cn may be depicted as follows:
a
b
c
d
By Lemma 19, a( a, c} = n-5, a(a,d) = n-6, a( b, c} = n-6 and a( b, d} = n-6 so
that 7(s4 ) = 4n-23. Note that this case can be simplified further. Since d,., (a, c}
and n 2 7, we have den (a, d}
=
3, de,, (b, c}
=
3 and d0, (b, d)
=
3 if n
=
=
2
7 while
de,. ( b, d} = 4 if n 2 8. Thus, for n = 7, we find 7(s4 } = 5.
Case 6. d,,. ( a, c) > 2, d,,, (a, d} > 2, d,,. ( b, c) > 2 and d,,.( b, d}
4(n - 6) = 4n - 24. •
a
b
c
d
>2
Note that n 2 8. By Lemma 19, a(a,c) = a( a,d} = a( b,c) = a(b,d} = n - 6 so
that 7(s4)
=
50
The following lemma gives the number of 3-cycles in sn which have an edge in
common with c4, where c4 is an induced chordless 4-cycle corresponding to a vertex
C4 in C(s.).
Lemma 27 Let c,1 = acbda be an arbitrary induced chordless 4-cycle in sn, n >
6, and let Sn = Cn· Then the number of 3-cycles Cs in
common with c4 is
2,
-y' (cl ) =
5
Sn
which have an edge in
ifdc.. (a,c) = 2, de,. (a,d) = 3, de,. (b,c) = 3, dc,. (b,d) = 2,
if dc,. (a, c)
= 2,
d,,. (a, d) = 3, d,.. (b, c)
n=G
= 3,
d,Jb, d) = 3,
4n - 23, ifdc.. (a,c) = 2, d0, (a,d) = 3, de.. (b,c) = 3, d,.. (b,d) = 4,
n=7
TI. e'. 8
4n - 24, ifdc.. (a, c) > 2, d,,. (a,d) > 2, d,.. (b,c) > 2, d,.. (b,d) > 2,
n>S
Proo£ sn = en by Theorem 13. T heorem 1 3 also implies that c4_
acbda in
corresponds to s,1 = { ab, cd} in en. Furthermore, for every 3-cycle in
=
which
has an edge, say ac in common with c,1, there is an induced 3-pointed star in
Sn
Sn
which contains a and c as non-adjacent vertices, and vice-versa. Hence, the result
immediately follows from Lemma 26. •
<;i.
We next determine the number of 4-pointed stars in a chordless cycle that have two
non-adjacent vertices in common with a fixed 4-pointed star.
51
Lemma 28 Given an induced 4 p- ointed star s4 = {ab, cd} in Cn, n > 61 the number
6(sn
. of 4-pointed stars 84 which have two non-adjacent vertices in common with s4
is
2,
4
if dc.. (a, c) = 2, d.,,. (a,d) = 3, de,. (b,c) = 3, dc,. (b,d) = 2,
n=6
if de,. (a, c) = 2, d,. (a, d) = 3, d,. (b, c) = 3, de,. (b, d) = 3,
n=7
5,
if dc.. (a,c) = 2, d.,,. (a,d)
= 3,
d,. (b,c) = 3, dc,. (b,d) =.4,
n 2'. 8
5(s:)
=
6
if dc.. (a,c) = 3, d.,,. (a,d) = 4, d,,. (b,c) = 4, dc,. (b, d) = 3,
n=S
7, if [dc,. (a, c) = 3, d�, (a,d) = 4, dc,. (b,c) = 4, d.,,. (b,d) = 4,
n = 9]
or [d,,, (a, c) = 3, d,. (a, d) = 4, d,,, (b, c) = 4, d,.(b, d) = 5,
n > 10]
8, ifd,. (a, c) 2'. 4, d,,, (a,d) 2: 4, d,. (b,c) 2'. 4, d,. (b,d) 2'. 4,
n > 10
We will prove this lemma directly. not making use of the values given in Lemma 22,
as the former is simpler for this situation. A disadvantage of applying Lemma 22 is
a
b
shown in the following example: Suppose that c,, contains :) ; , then in counting
c
d
the number of s4 which have two non-adjacent vertices in common with s4, the 4pointed star s4 = {ab, d2} is counted twice: first in detennining the value of f3(a, d)
which is the number of S4 for which a, d E V(s4)nV(s4 ), and second in determining
the value of (](a, d) which is the number of s, for which b, d E V(s,)nV(s4). Taking
this situation into account is more tedious and laborious.
Proo[ We note that with reference to Lemma 22, the number of 4 -pointed stars s4
containing two non-adjacent vertices in common with s4 depends on the distances
52
of each pair of these vertices in
en; these distances are either 2,
3 or at least 4.
Let O{s-4J be the number of 4-pointed stars containing two non-adjacent vertices in
common with s4,.
As seen in the proof of Lemma 26, the only permissible case where there are at
least two pairs of non-adjacent vertices of s4 whose distance is 2, is that in which
d.,. (a,c) = 2, d,. (a,d) = 3, d,. (b, c) = 3 and d,. (b,d) = 2, n = 6. We label "6 as
a b
201
c
d
Then the required s4 are {a2, di} and { bi, c2} so that 5(s,) = 2.
In those cases where there is exactly one pair of non-adjacent vertices of s,4 whose
distance is 2, it follows from the proof of Lemma 26 that the only permissible cases
are those in which d,., (a,c) = 2, d�. (a,d) = 3, d,. (b,c) = 3 and d.,. (b, d) = 3,
n = 7 or d,.(a,c) = 2, d,. (a,d) = 3, d,. (b,c) = 3 and d.,. (b, d) = 4, n 2 8.
If dc, (a,c) = 2, dc, (a,d) = 3, dc, (b,c) = 3 and dc,(b, d) = 3, n. = 7, we label c7
as
b
a
3
c
d
2
and thus the required s, are {a3,d2}, {ab,d2}, {bl,cd}, {bl, c3} and {ab,d2} so
that 5(s,) = 5.
If d.,. (a,c) = 2, d,. (a,d) = 3, d,. (b,c) = 3 and d.,. (b,d) = 4, n 2 8, we label Cn
as
53
a
b
n-2
c
and thus the required s4 are {a n
2
d
n-5
2, d n - 5), {ab, d n - 5}, {bl, cd), {bl , c n - 2)
and {bl, d n - 5) so that 6(s4 ) = 5.
At this point, we have eliminated those cases for which at least one pair of non­
adjacent vertices of s4 has distance equal to 2 in Cn· From now on, we assume that
the values of d,,. (a, c), d,,. (a, d), d,,. (b, c) and d,,. (b, d) are either 3 or at least 4.
First, we consider the case when den (a, c) = 3 and de,, (a, d)
following situations:
a
a
b
b
(iii)
(ii)
(i)
d
c
c
u
=
3. We have the
a
b
c
d
d
(i) and (iii) imply that d,,. (b, d) = 2 while (ii) implies that either c,, is not a cycle or
Cn has a chord. Both of these are contradictions to the assumption so that de,, (a, c)
and de,, (a, d) cannot be both equal to 3.
Now, consider the case where de,., (a, c) = 3 and de,, (b1 d) = 3. Then we have the
following possibilities:
a
a
b
b
or
(i)
c
d
a
d
b
c
d
a
b
or
(iii)
(ii)
c
a
b
c
d
c
d
54
(i) implies that do,, (a, d) = 2 while (iii) implies that d,.(b, c) = 2 which are both
contradictions to the assumption. On the other hand, (ii) satisfies all the assumptions.
Hence, whenever, do,, (a, c) = 3 and d,.. (b,d) = 3, then n = 8, do,, (a, d) = 4 and
do,, (b, c) = 4.
a
b
c
d
Then the required s4 are { a4, cd}, {a4, d2}, {ab, c3}, ( ab, d2}, {bl, cd} and {bl, c3}
so that o(s;) = 6.
Hence, the only pennissible case where at least two pairs of non-adjacent vertices
of s4 have distance equal to 3 in Cn ) is that in which dc,.(a,c)
d,.. (b,c)
=
=
3, de,, (a,d) = 4,
4 and d,..(b,d) = 3. So then we are left with two cases: dc,, (a,c) = 3,
d"' (a, d) 2 4, d,.. (b, c) 2 4, d,,, (b, d) 2 4; and d,.. (a, c) 2 4, d�, (a, d) 2 4,
d,.. (b, c) 2 4, d,,, (b, d) 2 4.
Suppose that do,, (a, c) = 3, d,.. ( a, d) 2 4, d,,, (b, c) 2 4, and d,,, (b, d) 2 4.
a b
d
Clearly n 2 9 and note that this case can be simplified further. Since dcJa, c) = 3,
we have d"' (a,d) = 4, d,.. (b,c) = 4 and d,.(b,d) = 4 if n = 9 while do,, (b, "' = 5
if n 2 10. We label c,, as follows:
55
b
a
n-2
2
n-3
3
c
d
n-6
Then the required s4 are {a n - 2,cd), {a n - 2,d n - 6) , {ab, c n - 3}, {ab,d n - 6} ,
{bl, c n - 3}, {bl, cd} and {bl, d n - 6} so that o(s,) = 7.
Finally, suppose that dc,, (a,c) 2 4, d,.. (a,d) 2 4, d,,. (b,c) 2 4, and d,.. (b,d) 2 4,
then en is
a
b
c
d
Note that n 2 10. We label Cn as follows:
.
a
b
n-2 � .1
k+5 '.,,. k
k+4�+ 1
c d
Then the required s4 are {a n - 2, cd}, {a n - 2, c k + 4), {ab,c k + 4}, {a n - 2,
d k+ l}, {ab,d k + l), {bl, cd}, {bl, c k + 4} and {bl, d k + l} so that 6(s4 )
8. •
=
56
Lemma 29 gives the number of induced chordless 4-cycles c4 -=/: c,4 in sn which
have an edge in common with c.;i, where c,1 is an induced chordless 4-cycle in sn
corresponding to a vertex 04 in C(sn), n 2: 6.
Lemma 29 l.£t c4
2:
61 and let sn = Cn· Then the number fl (c,4) of induced chordless 4-cycles C4 =/:- c4 in
=
acbda be an arbitrary induced chordless 4-cycle in sn , n
sn which have an edge in common with c,4 is
o'(c,) =
2,
if <l,. (a,c) = 2, d,,. (a, d) = 3, d,,. (b, c) = 3, d,,.(b, d) = 2,
4
ifd,. (a,c) = Z, d"" (a,d) = 3 , d,,, (b, c) = 3, d,.(b,d) = 3,
n=6
5 , if d,. (a,c) = 2, d,,, (a,d) = 3 , d,,, (b,c) = 3,d"" (b,d) = 4,
n=7
n>S
6
if d,. (a, c) = 3, d,,, (a,d) = 4, d,,, (b, c) = 4,d,,, (b,d) = 3,
n= 8
n = 9]
7 , if [d," (a,c) = 3, d,,, (a,d) = 4, d�, (b, c) = 4, d,,, (b, d) = 4,
or [d,.(a, c) = 3, d,,, (a, d) = 4, d,,, (b, c) = 4, d,,, (b, d) = 5,
n 2: 10]
Proo£
8, ifd,. (a, c) 2: 4, d,,, (a,d) 2'. 4, d,. (b, c) 2: 4, d,. (b,d)
=
n
2:
10
2: 4,
by Theorem 1 3 . Theorem 1 3 also implies that c.;i = acbda in
corresponds to s; = { ab, cd} in c;,. Furthermore, for every induced chordless 4-cycle
Sn
Cn
Sn
c4 f=. c;i in sn which has an edge, say ac in common with c4, there is an induced
4-pointed star in Cn which contains a and c as non-adjacent vertices, and vice-versa.
Hence, the result immediately follows from Lemma 28. •
We combine the results of Lemmas 27 and 29 to get degc{sn ) C,i , where O,i is a
vertex in 0( sn) corresponding to an induced chordless 4-cycle c4 in Sn ,
57
Theorem 30 Let C4 be a vertex in C(s,,) corresponding to an induced chord/ess
4-cycle c4 = acbda in Sn, n 2: 6. If Sn = Cn, then degC(s,. ) 0,1 has the value
4,
9,
if dc.. (a, c) = 2, d,. (a,d) = 3, dc,, (b,c) = 3, dc,, (b,d) = 2, n = 6
ifdc.. (a,c) = 2, d,. (a,d) = 3, dc,, (b,c) = 3, d.,. (b,d) = 3, n = 7
4n- 18,
ifJ.,.(a, c) = 2,
de. (a, d) = 3,
d.,. (b, c) = 3 , d.,. (b,d) = 4, n 2 8
4n - 17, if d.,. (a, c) = 3, d'" (a, d) = 4, de,, (b, c)
14,
if dc,, (a, c) = 3, d,. (a, d) = 4, de,, (b,c) = 4, d.,. (b, d) = 3, n = 8
= 4,
d,. (b, d)
= 4, 5, n 2 9
4n - 16, ifdc,, (a, c) > 4, d,. (a, d) > 4, d,. (b,c) 2 4, d,. (b,d) 2 4, n > 10
Proo£ It follows from Lemma 27 and Lemma 29 that degc(s,. )
of 3-cycles c3 in
which have an edge in common with c4 + the number of
induced chordless 4-cycles c4 =I c4_ in
c4
=
Sn
c;
=
the number
which have an edge in common with
1'(c4) + 8'(cD, Combining their values given in these two lemmas, we get
Sn
the desired r esult. •
C( sn), n 2: 6, one example for each of the two major cases considered.
We illustrate the previous results obtained concerning the degree of the vertices in
Let G* be a vertex in G( ss). We label ss as follows:
a,
Example 6 C* corresponds to the 3-cycle cj = a1a4a7a1
I.etting ss
= cs,
we find that dca (a1 1 a4)
By Theorem 25, we have degc(ss) G*
= 3, dc8 (a1, a1)
=
=
2 and dca (a4 1 a1)
=
3.
11. We check this value by looking at
s8 and counting the number of induced chordless cycles which have an edge in
common with cj. The required chordless cycles are a1a4a6a1, a1a3a1a1, a1a5a1a1,
58
Example 7 G* corresponds to the induced chordless 4-cycle c4_ = a1a5�ct(3a1
We find that dca (a,,a,)
=
4, dc,(a,,a,)
=
3, dc, (a2 , a5)
=
3 and dc, (a2 , a6)
=
4.
By Theorem 30, we have deg0(8s) G* = 14. To check: The induced chordless cycles
which have an edge in common with c4 are a1a3a5a1, a1a3aaa1, a1a4a5a1, a1a5a7a1,
Finally, we determine the order of C(sn) , n
2:
6, ie IV(C(sn ))I. We note that
V(C(sn )) corresponds to the set of induced chordless k-cycles, 3 :; k :; n in •n ·
But by Theorem 18, k i s either 3 or 4. Thus, IV(C(sn))I
the number of chordless 3-cycles in
Sn
=
n; + n;, where n; is
and n� is the number of chordless 4-cycles
in Sn,
Lemma 31 The total number ofpairs of non-adjacent vertices in Cn, n 2: 6 is n(n2-3) .
Proo£ Using the definition of the complement of a graph, the number of pairs of
non-adjacent vertices in Cn is equal to the number of edges in Cn· But the number
of edges in c,, is equal to the number of edges in the complete graph of order n number of edges in Cn · Thus the result follows. •
.
nn
Lemma 32 The number n3 of induced 3-pointed stars in Cn, n 2::: 6 is n3 = ( -4J(n-5)
Proo£ An arbitrary 3-pointed star { x, y, z} corresponds to three pairs of non-adjacent
vertices in c,, : {x,y), {x,z) and (y,z). Hence,
n3 =
L
{u,v}, u""v
a(u, v) / 3 -
(n - 5)
{u,v}, dc,. (u,11)=2
+
(n - 6)] / 3
{u,v}, d0,.(u,t1}>2
To determine the numbe r of pairs of non-adjacent vertices {u, v} in Cn, where
59
d.,. (u, v) = 2, let c,. = 123. ..nl. Then the required pairs of non-adjacent vertices are
{ 1, 3}, {2,4}, {3, 5} ,... {n - 2, n}, {n - 1, 1), {n, 2} so that there are n of them.
The number of pairs of non-adjacent vertices {u, v } in Cn where den (u,v) > 2 is
_ n(n-5)
n(n-3) _
n2
2
•
+!ili!..=fil
Hence, na = n(n-5) 3 (n-6) = n(n-4)G(n-5) • •
Proposition 33 The number n; of 3-cycles in sn, n 2:: 6 is n; = n(n-�(n-5) .
Proo£ We know that the chordless 3-cycles in Sn correspond to the induced 3-pointed
stars in en, n 2::: 6. Then by Lemma 32, we have n� = n3 = n(n-4J(n-5) . •
Our next goal is to detennine the number of induced chordless 4-cycles in Sn , n 2:: 6.
Lemma 34 The number n4 of induced 4-pointed stars in Cn, n 2:: 6 is n4 = n(n2-5)
Proo£ We first consider the case where n = 6. Let c6 = 1234561, then the induced
4-sided stars are {12,45}, {23,56} and {34, 16} so that n.4 = 3.
Assume that n
2:: 7. Note that every 4-sided star, say s4 = {wx, yz} corresponds to
four pairs of non-adjacent vertices in Cn : {w 1 y}, {w 1 z}, {x 1 y} and {x,z } in this
case. Hence,
n4 =
I;
{u,tJ}, u""tJ
f!(u,v) / 4 -
{u,tJ}, dcn (u,1!)=2
1+
3+
{u,v}, den (u,v);:;:4
4J I 4
From the proof of Lemma 32, the number of pairs of non-adjacent vertices { u, v} in
Cn where de,, (u1 v) = 2 is n.
Similarly, the number of pairs of non-adjacent vertices {u, v } in Cn where de,. (u, v) =
3 is n (The required pairs of non-adjacent vertices are {1, 4}, {2, 5}, {3, 6}, ... {n 3 , n),
{n - 2, 1}, {n - 1,2} and {n,3}).
The number of pairs of non-adjacent vertices {u, v} in 41. where den (u, v)
then n(n-3) - 2n = n(n4-7) by Lemma 3 1 and the two results above.
2
2: 4 is
60
- n(n-5)
Hence, n2 = (nxl) + (nx3)4+ <:¥1 x4) _
2
Note that if n = 6, n(n2-5) = 3 so that this formula also holds for n = 6. •
Proposition 35 The number n� of induced chordless 4-cycles in sn, n
n(n-5)
2: 6 is n�
Proo£ Since the induced chordless 4-cycles in Sn correspond to the induced 4-sided
stars in Cn, n 2: 6, we have r{4 = n4 = n(n;-5) by Lemma 34. •
Theorem 36 The order oJ C(sn) is n(n- lJ(n-5), n 2: 6.
Proo£ By Proposition 33 and Proposition 35, the order IV(C(sn))I of C(sn), n 2: 6
is n� + n� = n(n-lJ(n-5) . •
61
Chapter 4
On the Mapping R : G ---+ C ( G)
The definition of isomorphic graphs implies that two graphs are isomorphic if they
can be reconstructed in such a way that the resulting graphs have exactly the same
structure and differ only in the labelling of their vertices. In other words, isomorphism
expresses what, in less formal language, we mean when we say that two graphs are
"the same" graph.
In this dissertation, when we name the mapping G - C(G) R, ie R : G - C(G},
we then do not make any distinction between R(G) � C(G) and R(G) = C(G).
4.1 Properties of R
The purpose of this section is to exhibit a number of basic properties of the mapping
R : G - C(G).
Proposition 37 The domain of R is the set of all graphs while its range consists of
all cycle graphs and 0.
Proof. The complement of a graph is another graph while the cycle graph of a graph
can be another graph or 0. Thus, for any graph G, R(G)
=
C(G) always exists.
Therefore, Domain R = the set of all graphs.
graph !(, there is a unique graph G which is the complement of K, ie G = K, or
If H is a cycle graph, then there exists a graph K such that C(K) � H. For the
equivalently, G = K. So we see that if H is a cycle graph, there exists a graph G
such that R(G)
=
C( G) � H. Also, we have previously seen some graphs whose
62
cycle graphs of their complements are 0. Therefore, Range R = the set of all cycle
graphs U 0. •
Proposition 38 R is not one-to-one and is not onto.
Proot: First, we prove that R is not one-to-one. We need to show that there are two
graphs G1 and G, for which R(G1 ) '.:'. R(G,), ie C(G 1 ) � C(G,), but G1 � G2 •
Let G1 and G, be the following graphs:
l
2
G, : �
• 2
G 1 : /.
4
__.,5
3
l
For G 1 , the elements of V(R(G1 )) are
3l I - { { !, 2, 3}, { l, 2}, {l, 3}, {2, 3 } ) and
3
sj'I . . { {1, 3, 4} , { l , 3}, {l, 4} , {3, 4}).
And R(G,) = C(G1) is
s l(3)
s�
2
For G2 , the elements of V(R(G2)) are
sj'i -
{{l',3', 5'}, ,l', 3'} , {l', 5'} . . {3',5'}) and
sj'I
{{1'2', 4'5'}, {l', 4'}, {l', 5'}, {2', 4'}, {2', 5'}}.
And R(G2 ) = C(G,) is
63
s
(3)
l
s�
l
So we see that R(G1 ) � R(G2), but G1 ,>g G2•
Therefore, R is not one-to-one.
Now, we prove that R is not onto. We need to exhibit a graph G for which there
is no graph H whose cycle graph of the complement is G, ie R(H) = C(H) � G.
Consider a graph G for which C3 (G) # © and C4 (G) = ©. An example of such a
graph is shown below:
e
b
G:
h
c
a
g
We denote the cycles of G by 1 = abcda, 2 = bcdfeb, 3 = efihe, 4 = dfgd,
and 5 = abefda, then C( G) is
2
3
4
5
Let x = 1251, y = 2452, z = 2532, then C2 (G) is
y
z
64
Let A = xyzx, then C3 (G) is the single vertex A and G'(G) = 0.
Then by Theorem 7, there is no graph L for which C(L) � G. This means that there
can be no graph H for which R(H) = C(H) � G. Hence, R is not onto. •
As a direct consequence of Proposition 38, we have the following corollary:
Corollary 39 R has no inverse.
Proot: By definition, for a mapping to have an inverse, it needs to be one-to-one
and onto. Thus, by Proposition 38, R has no inverse. •
A graph without a cycle is called a forest (Harary, 1969).
At this point, we give a characterisation of the complement of a forest.
Theorem 40 A graph G is the complement of a forest if and only if G does not
contain any k-pointed star, k 2: 3.
Proot Let G be a graph of order n.
For n = 1 ) 2, the result is trivial. Corresponding to G, there exist,; a forest F such
that G = F, and G can never contain a k-pointed star sk since lski = k 2: 3
Assume that n 2: 3. We first show necessity:
Assume that
> n.
G is the complement of a forest, ie G is a forest. Suppose, to the
contrary, that G contains a k-pointed star sk, k � 3. By Theorem 13, sk is a chordless
k-cycle. G contains sk so that G contains a cycle. This is a contradiction since G is
a forest. Therefore,
G does not contain a k-pointed star, k � 3.
Next, we show sufficiency:
Assume that G does not contain any k-pointed star, k > 3. Suppose, to the contrary,
that G is not a forest. Then, by the definition of a forest,
G contains a cycle say Cm,
65
m ;:: 3. If Cm has a chord, there is a cycle of smaller length containing the chord
and parts of c,.: Suppose that ij is a chord of Cm
=
1 2...i i+l...j-1 j...m l, then
Cn = 1 2...i j...m 1 is a smaller cycle, ie n < m. Continue this process to generate
a chordless k-cycle
Ck,
for some k, 3 ::; k ::; n < m. Then we have shown that G
contains a chordless k-cycle ck, for some k > 3. By Theorem 13, the complement
of ck is a k-pointed star sk. And since G and G are complementary, G contains
sk,
This is a contradiction to the assumption. Therefore, G is a forest, ie G is the
complement of a forest. •
The corollary to this theorem given below tells us that R is a zero mapping under
the class of graphs not containing any k-pointed star, k � 3.
Corollary 41 R(G) = 0
if and only if G is a graph not containing any k-pointed
star, k 2'. 3.
Proof. By the definition of the mapping R, R(G) = 0 if and only if C(G) = 0. This
means that G has no cycle, ie G is a forest. Or equivalently, G is the complement
of a forest. Thus, the result follows immediately from Theorem 40. •
Next, we give two simple properties of the mapping R which associate R with other
mappings on graphs. We require the following definition of a self-complementary
graph (Harary, 1969):
A graph G is self-complementary if and only if G = G.
Proposition 42
( 1)
R is the composition of C, the cycle graph of the graph, and the function
mapping a graph to its complement.
(2)
R and C are equal Junctions under the class of self-complementary graphs.
Proof. (I) Let h be the function mapping a graph to its complement. Thus, for any
graph G, R(G) = C(G) = G(h(G)) = (G o h)(G)
66
(2) If G is a self-complementary graph, then G = G and so R(G) = C(G) C(G).•
Remark 2 It it easy to see that C a h =f:. h o G. As a counter-example, consider the
graph K(2, 3) below:
(C o h)(K(2, 3))
=
R(K(2, 3)). waking at the figure, V(R(K(2, 3))) consists of
one element sl3)
=
{a, b,c} so that (C o h){K(2, 3)) � K1 • On the other hand,
(h o C)(K(2, 3))
=
h(C(K(2, 3)))
=
h(I<,)
N,.
=
K3 =
N.1 ,
And we know that !{1
�
Finally, we enumerate some properties of an arbitrary graph under the mapping R.
Proposition 43 If H is an induced subgraph of a graph G, then R(H) is an induced
subgraph of R(G).
Proof. Since H is an induced subgraph of G, every induced star subgraph s,. of H is
in G and two vertices in
sk
are adjacent if and only if lhey are adjacent in G. This
implies that R(H) is a subgraph of R(G).
Now, Jet s(e), st•) E V(R(H)). Suppose that S{l)S(k) E E(R(G)), then we need to
show that Sl')S(k) E E(R(H)).
By assumption, 3(£) and so�) correspond to induced star subgraphs se and s,. in H.
respectively, which are then also in G. If 3(e) ,...., 3(k) in G, then the two stars have
two non-adjacent vertices, say u, v in common in G. Since these stars are in H, the
vertices u and v are also in H. Furthermore, u ri.J v in H since E(H) � E(G). This
implies that Sl') � S(k) in H. Therefore, the result follows. •
67
The next theorem tells us that the regular graph of a graph is the same as its regular
graph when some of its vertices of degree at least two less than its order, are deleted.
Theorem 44 Let G be a graph of order n, n <'. 3 and u E V(G). lfdega u = n - 2
or n - 1, then R(G) = R(G - u).
Proo[ We know that any k-pointed star Bk of G, 3 S k S n is (k - 3)-regular of
order k. Thus, every vertex of G which is contained in an induced k-pointed star
is of degree ::; n - 3. By assumption, dega u
>
n - 3 so that u is not a vertex
of any induced k-pointed star of G. This means that G and G - u have the same
set of induced k-pointed stars and hence, V(R(G))
=
V(R(G - u)). Also, deleting
u from G does not affect the structure of the induced k-pointed stars of G so that
E(R(G)) = E(R(G - u)). Therefore, R(G) = R(G - u). •
Remark 3 Alternatively, Theorem 44 can be proven by looking at G and G - u and
applying the definition of the mapping C :
If dega u
hand,
=
n - 1, then degau = 0 so that u is an isolated vertex in G. On the other
if dega u = n - 2, then degau = 1 so that if u
rv
v in G1 then UV is an edge
of G not in any of its cycle. Then deleting u from G does not affect the structure of
the chordless cycles in G and therefore, C(G) = C(G - u), ie R(G) = R(G - u).
This theorem can be extended to the following corollary:
Corollary 45 Let G be a graph of order n, n <'. 3 and H = {u E V(G)I dega u =
n - 2 or n - 1}. Then R(G) = R(G\H).
4.2 Identification of the Images Under R of Some
Special Graphs
In this section, we study the images under the mapping R of some special graphs.
These graphs include J("' K(m, n), Sm Cn, Wn and Pm as well as graphs obtained
68
by performing some finite operations on graphs.
We begin this section with some auxiliary results concerning the disjoint union and
the join of two disjoint graphs.
Let G, and G, be two disjoint graphs. Their disjoint union (Harary, 1969), denoted
by G , U G, , is the graph with vertex set V(G1 U G2) : V(G1 ) U V(G2 ) and edge
set E(G1 U G2 ) : E(G,) LJ E(G2). Their join (Harary, 1969), denoted by G1 +G,,
consists of G1 LI G2 together with all the edges joining all vertices of G1 to all
vertices of G 2 .
Lemma 46 For any two disjoint graphs G 1 andG2, G1 + G2 =G 1 UG2 . Similarly,
G, uG2 =G 1 + G,.
Proo£ Since G 1 and G2 are disjoint, we have V(G1 ) n V(G2 )
E(G2 )
= ©.
=
0 and E(G1 ) n
G1 (J G21 ie the complement of G1 + G2 is
G1 CJ G2• By the definition of the complement of a graph, we need to show that
Wt: wish to show that G1 +G2
V(G, + G,)
=
V(G, U G2 ) and that two vertices of G, + G, are adjacent if and
only if they are non-adjacent in G 1 U G2 ,
=
By definition,
G 1 + G2 = G1 U G,
Thus,
V(G1 + G,)
and E(G1 + G2 )
u {nvju E V(G,) and v E V(G,) }
V(G, ) (J V(G2)
E(G,) LI E(G2) (J {uvju E V(G1 ) and v E V(G2 ) }
On the other hand, G 1 U G2 is the graph where
V(G, u G,)
=
V(G, ) u V(G,) = V(G1 ) (J V(G,)
69
and E(G1 LI G2 )
-
E (G,) LI E(G2 )
- { uvlu, v E V(G,), u "" v in G, } LI {uvlu, v E V(G,), u "" v in G,}
Hence, we have shown that
V(G, + G,) = V(G,) LI V(G,) = V(G, LI G,)
To show that the edge sets of G1 + G2 and G1 U G2 coincide, it suffices to show
that two vertices in G1 + G2 are adjacent if and only if they are non-adjacent in
G 1 LI G,.
Since V(G1 ) n V(G2)
=
0, we consider only the cases where either two vertices x
and y are in the same vertex set V(Gi) , i = 11 2 .or one vertex x is in V(G1 ) while
the other vertex y is in V(G,).
Case I. x, y E V(Gi ) for exactly one i, i = 1, 2.
If x is adjacent to y in G;, then xy E E(G;) ,;; E(G1 + G2 ) . Thus, xy E E(G1 + G,)
which means that x ,.,,, y in G1 + G2 • On the other hand, x """ y in Gi · But Gi is a
subgraph of G1 U G2 so that x rp y in G1 (J G2 • So we have x rv y in G1 + G2 while
x 17"' y in G1 (J G2 .
If x and y are non-adjacent in G;, then xy
V(G;) for exactly one i, xy
rt
rt
E(G1 ), E(G,). Also, since x,y E
{uvlu E V(G,) and v E V(G,)} . Then xy
E(G1 ) LI E(G2 ) U {uvlu E V(G1 ) and v E V(G,)}
=
rt
E(G, + G,) so that x "" y
in G1 + G2 . On the other hand, x ,...., y in Gi, so that x ,...., y in G1 U G2, So we have
x """ y in G1 + G2 while x ,...., y in G1 U G2.
Case 2. x E V(G1 ) and y E V(G2 )
Then xy E {uvlu E V(G,) and 11 E V(G,)} ,;; E(G, +G,) so that xy E E(G1 +G,)
which means that x � y in G1 +G2 • On the other hand, xy rt E(G1 ) since y rt V(G1 )
and xy
'le E(i'J;")
since x
rt
V(G,). Hence, xy
'le E(G,) LI E(G,j = E(G1 LI G,j .
70
T his means that x ,.,., y in G1 (JG2 • So we have x "" y in G 1 +G2 while x r/J y in
G1 (J G,.
We have thus shown that two vertices in G1 + G2 are adjacent if and only if they
are non -adj acent in G1 U G2. Therefore, G1 + G2 = G1 U G2.
To show the second part of the lemma, we apply the first result to the graphs G1 and
G2 to get G1 + G2 =
(J G, = G1 (J G2. This means that G1 + G2 and G1 (J G,
C,
are complementary, ie G1 U G2 =G1 +G2, •
Theorem 47 For any two disjoint graphs G1 and G,, R(G1 + G,)
R( G1) U
R(G,).
Proof. First, we note that for any two disj oint graphs G and H, G(G (J H)
=
G(G) 1j G(H) since the cycles of G and the cycles of Hate independent from each
other, as their vertex and edge sets are disjoint.
Now, R(G 1 +G2 )
=
C( G 1 +G2)
= C(G1 (J G2) by Lemma 46.
above argument, it follows that R(G 1 +G2 ) = C(G1 ) (J C(G,)
=
And so with the
R(G1) U R(G,). •
In the succeeding discussions, we will be encountering graphs which result from
either deleting some vertices, deleting some edges or adding some edges from/to
a given graph. In certain instances, these graphs are of great interest since they
retain and lose some specific properties of the original graph. We introduce the
corresponding graph operations here.
The removal of a vertex u from a graph G results in the subgraph G - u of G
consisting of all vertices of G except u and all edges not incident with u. On the
other hand, the removal of an edge e from G yields the spanning subgraph G - e
of G containing all edges of G except c. The removal of a set of vertices or edges
from G is defined by the successive removal of single elements of the set (Harary,
1969).
71
If the vertices u and v are not adjacent in G, the addition of edge e = uv results in
the s mallest graph G + e containing G and e (Harary, 1969).
We now look at the regular graphs of some simple and common types of graphs.
Proposition 48
(I)
The regular graph R(Nn ) af a null graph Nn of order n, n 2 4, is G(n, 3,2).
(2)
The regular graph R(Kn ) of a complete graph Kn of order n is 0.
(3)
The regular graph R(K(m, n)) ofa complete bipartite graph K(m, n), m, n 2
4 , is G(m, 3,2) (J G(n, 3, 2).
(4)
(5)
The regular graph R(Sn) af a star graph Sn af order n, n 2 4, is G(n, 3, 2).
The regular graph R (c,,) ofa chordless cycle Cn oforder n, n > 3, is C(sn) ,
(6)
The regular graph R(Wn ) of a wheel Wn of order n, n 2 3, is C{sn)·
(7)
The regular graph R(Pn) of a path Pn of order n, n 2 3, is C(sn + e).
Proo£
(I) We note that Nn = Kn so that R(Nn) = C(Nn ) = C(Kn )· Hence, with Theorem
8 of Section 1.5, we have R(Nn ) = G(n, 3, 2), n 2 4.
(2) Equivalent to Nn
= Kn
is Kn
=
Nn so that R(Kn )
=
C( Kn )
=
C(Nn ),
But a nuli graph has no edges and hence has no cycle. Therefore, it follows that
R(Kn ) = 0.
(3) We note that K(m,n)
=
Nm + Nn so that R(K(m,n))
=
R(Nm + Nn),
Thus, by Theorem 47 and by (I), we have R(K(m, n)) = R(Nm ) (J R(Nn ) =
G(m, 3, 2) (J G(n, 3, 2), m,n 2 4.
(4) By the definition of Sn and by (3), R(Sn) = R(K(l,n)) = G{l, 3, 2) (J G(n, 3, 2),
n ;:: 4. And since G(l, 3, 2) = 0 by the definition of the uniform intersection graph,
we have R(Sn ) = G(n, 3, 2), n 2 4.
(5) By Theorem 13, we have R( c,,) = C{c,,) = C{sn), n > 3.
72
(6) By the definition of Wn and by Theorem 47, we have R( Wn) = R( K1 + c,,) =
R( K,) U R( c,,), n 2: 3. But by (2) and (5), R( K,) = 0 and R( c,,) = C( sn ), n 2: 3
SO
that R( Wn) = 0 U C( sn) = C( sn ), n 2 3.
(7) We note that Pn = c,, - e, for some edge e of c,,, n > 3 so that R( Pn ) =
C( Pn ) = C( c,, - e), n 2: 3. By Theorem 13, c,, = sn; so then, Cn
e = Sn + e.
Thus, i t follows that R( Pn) = C( sn + e), n 2: 3. •
We next discuss the regular graph of a tree. As the complements of some trees are
flip-flap wheels, we first determine the cycle graph of a wheel and show that flip -flap
wheels are cycle periodic. These results are found i n Egawa, Kano & Tan (1991);
however, their proofs are not given there. Since these lemmas are necessary in the
succeeding discussion, we present their proofs here.
Lemma 49 For n 2: 3, C( Wn ) � Wn 1(1 + Cn, n > 3. We Jet
Proo£ By definition, Wn
1(1
be the single u and
c,, = 123...n.
n
2
·
.�3
. u
,_ . - -
The chordless cycles in Wn are
a1
= lu21 1 a2 = 2u32,
a3
= 3u43, ...,a71 = nuln
and v = 1234...nl. Then in C( Wn), ai ,...._, a;+1 and an ,...., a1 so that a1 a2 ...an a 1 forms
a chordless n cycle and v ,....., ai, for all i = 1, 2, ..., n. Thus, G( Wn ) is
a
on
,
a,
··
. �a
, V
3
·. - 04
Therefore, C ( Wn) � Wn for n 2: 3. •
� Wn
73
Lemma 50 Every flip-flap wheel with some flip orflap is cycle-periodic ofperiod 1
or 2.
Proo£ Let us consider a flip -flap wheel G containing a wheel Wn and some flip
c, n > 3.
By Lemma 49, we know that if a 1 = lu21 1 a2
and v
=
=
2u321 a3
=
3u43, ..., an = nuln
1234...nl, then C( Wn ) is a graph isomorphic to Wn. Now, the flip c
contains the edge lu which is common to exactly two chordless cycles of Wn,
namely, a1
=
lu21, and an = nuln. Thus, correspollding to the wheel Wn and the
flip c of G,C( G) contains the following graph which is a wheel Wn together with
the flap c* = ca1 anc.
c
edge a1 a,. is common to the flap c' and to x,. and w, so that C( C( G}}
=
C2 ( G}
contains the graph below which is a wheel Wn togethei with the flip Xnc"'wxri -
74
Thus, a flip in G corresponds to a flap in C(G) which corresponds to a flip again
in C2 (G). Similarly, a flap in G corresponds to a flip in C(G) which corresponds
to a flap again in C2 (G). Therefore, for a flip-flap wheel G with some flip or flap,
either C(G) � G or C2 (G) '.::'. G, ie G is periodic of period either I or 2. •
Proposition 48, # 4 and 7 give us the formulas for the regular graphs of a star and of
a path, respectively. As we know, stars and paths are types of trees. In general, there
is no specific fonnula for the regular graph of an arbitrary tree. However, depending
on the order n of a tree Tn , the classification of R(T,i) based on its cycle graph can be
detennined. It can be observed that there is a transition of R(Tn ) from a n�ll graph,
to a cycle-vanishing, to a cycle-periodic and finally to a cycle-expanding graph as
the number of vertices of the tree increases. We have the following proposition:
Proposition 51 Let Tn be a tree of order n. Then
(i)
(ii)
For n = 1 , 2, 3, R(Tn ) = 0.
For n = 4, either R(Tn)
(iii) For n = 5,
=
0 or R(Tn ) is cycle-vanishing.
R(Tn) is either cycle-vanishing or cycle-periodic.
(iv) For n = 6 1 R(Tn) is either cycle-periodic or cycle-expanding.
(v) For n > 6, R(Tn ) is cycle-expanding.
Proof. The proof for n � 6 is done by the enumeration process.
(i)
Clearly, for n = 1, 2, 3, Tn has no cycle so that R(Tn ) = C(Tn ) = 0.
(ii)
For n = 4, T4 is either P4 or 83 .
By Proposition 48, #7, R(P,)
R(P,) = C(P,) = 0.
We label 83 as follows:
=
C(s, + e). Since s, + e is isomorphic to P,,
75
s,, y
a
b
Thus R(S3 ) is the single vertex s;'l = {{a, b, d}, {a, b} , { a, d}, { b, d}) so that R(S3)
is cycle-vanishing.
(iii) For n = 5, T5 is either P,, S, or the graph
6,
I
By Proposition 48, #7, R(P5 ) = C(s5 + e). And since any s5 + e is isomorphic to
the graph
R(P5 ) '!a! P2 which is cycle-vanishing.
C(I(4) . But K4 - W, and so C(K4 )
By Proposition 48, #4, R(S4 )
=
G(4, 3 , 2). By Theorem 8 of Section 1.5, G(4, 3, 2) �
�
C(W3 ) = W3 by Lemma 49. Thus, we have
R(S4 ) � C(J{4 ) � W, which is cycle-periodic.
If T5 is the graph below:
a
b
R(T5) can be determined by its definition:
sl3> = { {a, b, d), {a, b), {a, d) , (b, d}}
s(3)
,
Then R(T5 ) is /
s (3)
2
'!a!
sf> = [{a, b, e}, {a, b}, {a, e} , {b,e}}
P2 which is cycle-vanishing.
76
(iv) For n = 6, Tn can be any one of the following graphs:
(a) P• : I I
I I
1
(d) .--,.::
1
(b) S, : -¥­
(c) I
I
I
(•)+
<
(0
We consider each of these 6 graphs.
>-<
(a) P•
Clearly, the complement of Pa is s, + e. Thus, R(P6 )
Theorem 16. sa is cycle-expanding. Since so + e contains so,
=
C(Pa)
expanding. And so is C(s6 + e). Hence, R(Pa) is cycle-expanding.
(b)
s,
Clearly, the complement of 85 is [(5 U N1 . Thus, R(S5 )
s5
C(sa + e) . By
+ e is also cycle­
=
C(S5)
C(K5 U N1 ) .
Since K5 U N1 contains 1(5 , by Theorem 1 of Section 1 .5, !{5 (J N1 is non-planar
=
=
and so is cycle expanding by Theorem 6. This means that C(I<5 U N1)
cycle-expanding, too.
(c) If T6 is isomorphic to � : �
<
a
b
R(S5 ) is
, we obtain R(Ts) by applying the
definition of R: The elements of V(R(T,)) are sl3l, sj'I,
sJ4) where
sj'I sj'I -
=
{{a,b,d}, {a, b), {a, d) , {b,d})
{ {a,b, e), {a, b), {a, e}, {b, e))
sj'I, sl'l, sj'I, sj'l and
sfl
si'I
S�'I
S;'1
s. 2<•I
-
77
{{ a, b, f}, { a, b}, { a, f), {b, .f)}
{ { a, d, f } , { a, d} , { a, f), {d, !) }
{ {b, d,f}, { b, d} , {b, !), { d, .f) }
- { { ac, ef } , {a, e}, { a, .f) , {c, e}, {c, .f} }
- {{ be, ef} , {b, e} , { b, .f}, {c, e), { c, .f}}
Then R(T,) is
s (4)1
Note that R(T6) contains the graph below which is isomorphic to J3 of Theorem 5
of Section 1.5 with s?)
x and sJBl -+ y:
----,
)
s<a
s<'4l -_....-rlf.N' s (4)
2
�:..-�s <a)
2
aJ
s<
By Theorem 5, R(T,) is cycle-expanding.
(d) If T, is isomorphic to f
a
b
, then To is the following graph:
d
78
a
c�:
b
We observe that T, is a flip-flap wheel with one flap. By Lemma 50, R(T,) = cm)
is a flip-flap wheel with one flip and is cycle -periodic.
a
(e) Since T6 is isomorphic to t � b
a
c
c
, T6 is the following graph:
b
We observe that T6 contains the graph below which is isomorphic to H3 of Theorem
5 of Section l.5 with f --> x.
a
�f
c
b
By Theorem 5, To is cycle-expanding and so is C(T6) = R(T,).
O
(!)
As T6 is isomorphic to dK
e
b
, then T6 is the following graph:
/·'
79
a
b
e
We observe that T6 is a flip-flap wheel with one flip and one flap. By Lemma
50, C(T,)
=
cycle-periodic.
R(To) is also a flip-flap wheel with one flip and one flap, and is
(v) n > 6.
We isolate the case where n
=
7. Up to isomorphism, there are eleven types of T1
including P1 and So,
(1) d e
I
I
a
f
I
g a c b
I
I
t
I
c
f e
(6) go-..
_,,-o--e b
a
g
e
°'\,.
d
a
d c
a
b
b
c
b
d
e
d
g
e
g
a
(7) --------
d
e d
(3) g. 1. , -• - s/
(4) f
f
e
(2) f * b
d
b
(9)
f
a
g
80
a
(5) d
e f
a
f
g
(10) * b
g
c
b
d
a
d
(11) �
/e · c\
b
f
Types 1-10 have the following properties: Vertices a,b,d and f are pairwise non­
adjacent; vertex e is not adjacent to vertices a, b and g; and vertex g is not adjacent to
vertices b, d and e. Thus, in their complements, a1 b1 d and f form a K4 � W3 , e rv
a, b, g and g ,..., b, d, e. This means that their complements contain the graph below
which is isomorphic to Ji of Theorem 5 of Section 1.5 with e --+ x and g -t y :
a
b
f
g
By Theorem 5, T7 of types 1-10 are cycle-expanding, and so is C(T1) = R(T1).
On the other hand, in T7 type 11, vertices a, b, d, f are pairwise non-adjacent and
g
f'JIJ
a, b, d. This means that its cdmplement T1 contains the graph below which is
isomorphic to H3 of Theorem 5 with g -+ x.
�g
f
b
By Theorem 5, T7 of type 1 1 is cycle-expanding, and so is C(T7) = R(T1 ).
Let n > 7. We wish to show that R(Tn ) is cycle-expanding. Firstly, we take note of
81
a trivial result on trees: (Harary, 1969)
For any tree Tn of order n, n
e'. 1,
IE(Tn)I = n - 1.
Now, we look at Tn . From the definition of the complement of a graph, we get
IV{Tn)I
= IV{Tn)I = n and
IE{Tn)I
= (;) - IE{Tn)I = (n-l�n-2).
We will prove that Tn is non-planar. As n > 7 implies that n - 1 > 6 and n - 2 > 0,
we have (n - l){n - 2) > 6{n - 2) or equivalenly, (n-l�n-2) > 3n - 6. Therefore,
Tn , n > 7 is non-planar by Corollary 3 of Section 1.5 and hence is cycle-expanding
by Theorem 8. This means that C(Tn) = R(Tn), n > 7 is cycle-expanding, too. •
Next, we perform some of the graph operations earlier discussed on the n-pointed
star and detennine the images of the resulting graphs under R and R2 .
Theorem 52 R(sn - u) = 0 for n e'. 3.
Proo£ We note that deleting a vertex in a graph does not affect the adjacency relation
of the other vertices with each other in the graph, nor the adjacency relation of these
vertices with each other in the complement of the graph. This means that for any
graph G and u E V{G), G - u = G - u.
Now, by the above argument and by Theorem 13, we have R(sn -u) = C(sn - u) =
C(sn - u) = C(Cn - u), n 2 3. Observe that Cn - u = Pn-1 and since a path has no
cycle, it follows that R(sn - u) = C(Pn- 1 ) = 0. •
Theorem 53 R(s. + e) = 0 for n 2 3.
Proo£ By Theorem 13, Cn = Sn, so then Sn + e = Cn - e, n 2: 3. Also, by definition,
Pn Cn - e for some edge e in Cn, n 2 3. Thus, we have R(sn + e) C(sn + e)
=
C(en - e)
=
C(Pn) = 0, n e'. 3. •
Theorem 54 R2 (sn LI K,) = C(sn), n e'. 3.
=
=
82
Proof For n 2 3, R2 (•n LI K,) = R(R(sn LI K1 ) = R(C(sn LI K1 )) = R(C(sn +
K1 )) by the second part of Lemma 46. But by Theorem 13 and by the definition of
Wn,
Sn
+ K1 = Cn + K1 = Wn, and so with Lemma 49 and Proposition 48, #6, we
get R2 (sn U K1 ) = R(C(Wn)) = R(Wn) = C(sn), n 2 3. •
Theorem 55 R2 ((s., LI {u} ) + (s,,, LI {v} )) = C((sn, U {u}) + (s,,, U {v} )), u f
Proo£ First, we note that the proof of Theorem 54 gives Sn U I<1 = Wn, or equiv­
part of Lemma 46, we obtain for u f v, n,, n2 2 3, R2((sn, LI {u) )+(sn, LI { v) )) =
alently, Sn U K1 = Wn , n > 3. Now, with Theorem 47, Lemma 49, and the second
R(R((sn, LI {u}) + (sn, LI {v}))) = R(R(sn, LI {u)) LI R(sn, LI {v})) =
R(C(sn, LI {u}) U C(sn, (J {v))) = R(C( Wn ,) LI C(Wn, ) ) = R(Wn, LI W., ) =
C(Wn, LI Wn,) = C(Wn, + vVn, ) = C((sn, LI {u}) + (sn, LI (v})). •
The next corollary is a special case of Theorem 55 while the second next one is an
extension of this theorem.
Corollary 56 R2 (!{(4,4)) = C(I<(4, ,J))
Proo£ By their definitions, we can deduce that K(4, 4)
= N,+N, and N, = s, LI K1,
And so with Theorem 55, we have for u f v, R2 (K(4,4)) = R'(N, + N,) =
R2 ((s3 LI {u}) + (s3 LI (v))) = C((s3 LI (u}) + (s, (J {v})) = C(N, + N, ) =
C(I<(4,4)). •
•
•
i=l
i=l
Corollary 57 R2 ( I; (sn, LI {a, })) = C(I; (sn, U {a,))), a, f a; for i f j, k 2
•
2 where E denotes the join of the k graphs
•
i=l
Proof. The proof is done by induction on k.
To finish this section, we look at the regular graph of the disjoint union of the
complete graph with some other graphs. We require the following lemma:
83
Lemma 58 The complete graph Kn does not contain any induced k-pointed star,
3 :,:; k :,; n.
Proot: Suppose, to the contrary, that Kn contains sk, 3 $ k $ n as an induced
subgraph. Then Kn contains Bk, By Theorem 13, sk is the chordless cycle ck. This is
a contradiction since Kn is the null graph Nn and hence does not contain any cycle.
Therefore, Kn does not contain an induced k-pointed star, 3 $ k $ n. •
Theorem 59
If G is the graph obtained byforming the disjoint union of Kn , n 2:: 1 ,
and a pair of non-adjacent vertices, then R(G) ......., Kn ,
Proo[ By assumption, G = KnU{a,b}. Let V(Kn ) = {1,2, ... , n}.
We isolate the cases where n = 1 1 2 .
If n = l, then G ::::: N3 and so R(G) is a single vertex which is isomorphic to K1 .
If n
=
2, then G is
2
1
1
I
a
I
{1, a} , {1,b}, {a,b)) and
b
3
.
And so V(R(G))) consists of s;'I
=
{{1, a,b},
sl I = {{2, a, b}, {2,a}, {2,b}, (a,b}) so that R(G) is
the edge s;3> sJ3) which is isomorphic to K2 ,
Let n <'. 3. By Lemma 58, I<n does not contain any k-pointed star, 3 :,:; k $ n
as an induced subgraph. Thus, the only k-pointed stars of G are the 3-pointed stars
of the form {i,a,b} 1 i
sP> =
=
1,2 1 ... 1 n. By Definition 2, the vertices of R(G) are
{{i, a,b}, {i,a} i {i,b}, {a,b}}, i
=
1 , 2, ... , n ) w hich are pairwise adjacent
since {a, b} is common to all i. Thus, R(G) is the graph consisting of n vertices
sf}, sfl, ... , 8�3) which are pairwise adjacent. This implies that R(G) "' Kn by the
definition of Kn. •
Given a graph G, the line graph L(G) of G is the graph whose vertices me the
edges in G and two vertices in L(G) are adjacent if and only if the corresponding
edges have a vertex in common (Harary, 1969).
84
Theorem 60 If G is the graph obtained by fanning the disjoint union of Kn, n 2:
1, and a single edge, then R(G) � L( I<n), n 2': 1.
Proof. By assumption, G = Kn (J {ab).
We isolate the cases where n = 1, 2.
•
If n = 1, then K, U {ab) is oo---e
(J {a,b)) = 0. On the other
o so that R(K1
a
b
hand, since E( I<,) = 0, we have L( I<,) = 0. Thus, R(K, U {ab)) � L(I<,).
•
•
a
b
If n = 2, then K2 U {ab }is ••---eo so that R( I{2 (J { ab}) is a single vertex. On the
other hand, since !{2 is a single edge, we have L( I<2)
�
J{1 . Thus, R(J{2 (J {ab)) :::
L(I<,).
Let n
k
2:
3. By Lemma 58, Kn does not contain any induced k-pointed star1 3 ::;
< n. Thus, for any sk of G, we have sk n ab f. 0 so that sk is disconnected.
The only disconnected k-poimed stars are s3 and s4 • However, G does not con­
tain any three non-adjacent vertices so it follows that the only k-pointed stars of
G are s4 , ie a pair of n011-adjacent edges. Since s4 n ab
i-
© and every two
edges in Kn are adjacent, s4 consists of ab and an edge in K11 • Note that ev­
ery s4 corresponds to a vertex in R(G) i ie if s�
=
{a.b,xy}, xy E E(I<n ), then
S;'l = { { ab, .,y), {a, x}, { a, y}, {b, x ) , {b, y ) ) E V(R(G)), so that there is a one-to­
one correspondence between the vertices of R(G) and the set of edges in Kn . Also,
two vertices in R(G) are adjacent whenever the two corresponding s,1 have a pair of
non-adjacent vertices in common. If sj'i = { { ab, xz}, {a, x}, {a, z}, {b, x), {b, z}},
then SJ4)
,....,
sj4) as {a,x} , {b,x} E s;4) , sj4). Jn this case, the edges xy and XZ in Kn
have x JS a common vertex. Thus, two vertices in R(G) are adjacent whenever the
two corresponding edges in Kn have a vertex in common. Therefore, R( G) � L(Kn)
by the definition of the line graph L( I<n) of I<n, •
85
4.3 Behaviour of Some Special Graphs Under Iter­
ates of R
We recall that a graph G is classified as either cycle-vanishing or cycle -persistent
depending on whether there exists a positive integer k for which Ck ( G) = 0, or there
is none. Analogously, for the mapping R, we introduce here R-vanishing and R­
persistent graphs.
Definition 4 A graph G is R-vanishing if and only if there exists a positive integer
k for which R'(G)
= 0.
Otherwise, G is said to be R-persistent.
An R-persistent graph G is R-expanding whenever kJim IF(R'(G))I
-oo
oo.
It can easily be seen that a complete graph of order n i s R-vanishing for n � 1 while
an n -p ointed star is R-vanishing for n 2: 3. We wish to determine the minimum order
of a null graph, a path, a cycle, a complete bipartite graph and a star for which they
are R-expanding. To be able to do this, we need to d i scuss some auxili ary results.
First, let us adopt the notation l> to indicate an induced subgraph relation, ie G l> H
means that H is an induced subgraph of G.
The lemma belows follows from Proposition 43 and mathematical induction:
Lemma 61 l.£t G and H be two graphs sttch that G I> H, then R'(G) !> Rk (H)
for k 2 1.
Proposition 62 let G C> H, then
f. If G is R-vanishing, then H is also R-vanishing.
2. If H is R-expanding, then G is also R-expanding.
86
Proo£ 1 . By assumption, there exists a positive integer k1 for which R•, ( G) = 0.
Also, by Lemma 61, Rk'(G) � R•• (H). Thus, we have 0 � Rk'(H). This meaos
that there is a positive integer k2 � k1 such that R"' (H)
·-00
2. By assumption, lim jV(R'(H))I
-
= 0,
ie H is R-vanishing.
oo. Also, by Lemma 61, Rk (G) �
R•(H) for an arbitrary k 2 1 so that V(R•(G)) 2 V(Rk(H)). Let k --> oo, then
·-00
IV(R'(H)) I --> oo and so !V(R'(G))I --> oo. This implies that lim I V(R•(G))I oo, ie G is R-expanding. •
Now, we determine the minimum order of a null graph for which it is R-expanding.
The lemma below is a result on the second iterated regular graph of a null graph of
order at least 6.
Lemma 63 R2 (Nn) � Nn+2 , n 2 6.
know, G(n , 3, 2) is the graph whose vertices are the subsets of cardinality 3 of the
Proo£ By Proposition 48,#1 , R2(Nn )
n-set S
=
=
R(R(Nn )) = R(G(n, 3, 2)), n 2 6 . As we
{1, 2, 3, ... , n}; and two vertices are adjacent in G(n1 3, 2) whenever the
corresponding two 3-subsets have exactly two elements in common.
To prove that for n 2 6, R(G(n, 3,2)) � Nn+2 , ie R(G(n, 3 , 2) ) contains (n + 2)
pairwise non-adjacent vertices, we will show that there are (n + 2} 3-pointed stars
in G(n, 3, 2) any two of which do not have two non-adjacent vertices in com mon.
For n = 6, noting thatthe vertices ofG( 6 , 3, 2) are the 3-subsets of S = {I, 2, 3, 4, 5, 6},
then the required 3-pointed stars are
{ { 1 , 2, 3}, {1,4,5}, {2,4,6} }
{ {2,3,4}, {2,5,6}, {3, 5, 1} }
{ {3,4,5}, {3, 6, l}, {4, 6,2 } }
{ {4,5,6}, {4, 1 , 2} , {5, 1 , 3} }
{ {5,6,1}, {5 ,2, 3}, {6,2,4} }
I
87
{ {6, 1, 2}, {6, 3, 4}, { 1, 3, 5} }
{ {1, 2,5}, {1, 3 , 6} , {2, 4, 5} }
{ {2,3,6}, {2, 4, l }, {3,5,6} }
Each of these sets is a 3-pointed star as any two elements of each stars are sets
which have at most one element in common. Now, these 8 3-pointed stars are
pairwise non-adjacent since any two of the sets intersect in at most one element.
For n = 7, the required 3-pointed stars are
{ {1, 2, 3}, {1, 4, 5}, {2, 4, 6} }
{ {2, 3, 4}, {2, 5, 6}, {3, 5 , 7} }
{ {3,4,5}, {3, 6,7}, {4, 6, 1} }
{ {4,5,6}, {4, 7,1}, {5, 7, 2} }
{ {5, 6, 7), {5, 1, 2}, {6, 1, 3} }
{ {6, 7, 1}, {6, 2, 3}, {7, 2, 4} }
{ {7, 1, 2}, {7, 3,4}, { 1 , 3, 5} }
{ {l, 2,4), {1, 3,5}, {3, 4, 6} }
For n � 8, we define
{ {2, 3, 5}, {2, 4, 6}, { 4, 5, 7} }
A 1 (k) - {(l + k) mod n, (2 + k) mod n, (3 + k) mod n )
A2 (k) - {(l + k) mod n,
A3 (k) - {(2 + k) mod n,
(4 + k) mod n,
(4 + k) mod n,
(5 + k) mod n}
(6 + k) mod n}
for k = 1 , 2, ... , n. Note that each of A1 (k) , A2 (k) , A,(k) can be regarded as a
set of 3 vertices in Cn = 1 23 ... nl such that if Ai(k) = {u, v , w} , then the set
{ d,,. (u, v) , d,,. (v , w) , d,)u, w)} of distances is unique for each i = 1 , 2, 3 : { 1, 1, 2}
corresponds to A,(k); { 1 , 3 , 4} to A,(k); and {2, 2 , 4} for A3 (k). Thus, these sets
have the following properties:
A,(k)
fa A,(!) ,
if l fa k
88
if i cl j
A;(k) cl A,(l),
I A;(k)
n A,(k)
I <
A,(k) n A,(k) n A,(k)
Thus, S(k)
S(k)
=
1
0
{A1 (k), A 2(k), A3 (k)} is a 3-pointed star in G(n, 3, 2). Moreove.,
n S(l) = 0 for k cl l, 1 < k, l < n so that the n sets S(k), k = 1, 2, ... , n form
n pairwise non-adjacent 3-pointed stars in G(n 1 3, 2). Finally, the 3-pointed stars
S(n + 1)
{ {1, 2, 4}, {1, 3, 5), {3, 4, 6} ) and
S(n + 2)
{ {2, 3, 5), {2,4, 6), {4, 5, 7) }
are disjoint and
(1 , 3, 5) if k = n - 1, i = l
S(k)
n S(n + i) =
(2,4,6)
if k = n, i = 2
0
else
Therefore, R(G(n, 3, 2)) I> Nn+2· •
We extend this lemma to obtain a result concerning the kth iterated regular graph of
a null graph of order at least 6, where k is an even positive integer.
Corollary 64 R2m (Nn) t> Nn+2m , m ?: 1, n 2: 6.
Proof. As a consequence of Proposition 43, we can successively apply R2 as an
operator to the relation R2(Nn ) !> Nn+2 given in Lemma 63 (t - 1) times, t ;:;: 1 to
get R"(Nn ) I> R2it- l)(Nn+2), n � 6. We thus obtain
Since I> is transitive, we get R2m (Nn ) !> Nn+2m , m ?: 1 1 n ?: 6. •
We can now prove that the minimum order of a null graph for which it is R-expanding
is 6.
89
Proposition 65 The null graph Nn of order n is R-vanishing for n < 6 and R­
expanding for n 2: 6.
Proo£ To show that Nn is R-vanishing for n < 6, it suffices to show by Proposition
62, #1 that Ns is R-vanishing since N5 t> Nn for n < 6.
We look at R(N5). By Proposition 48, #1, R(N5 ) "' G(5, 3, 2). If our 5-set is
S = {1, 2, 3, 4, 5}, then the vertices of R(N5 ) are a = {!, 2, 3}, b = {l, 2, 4),
c = {l, 2, 5 } , d = {1 , 3, 4), e = {l, 3, 5}, f = {!, 4, 5}, g = {2, 3, 4), h = {2,
3, 5}, i = {2, 4, 5} and j = {3, 4, 5) with the following non-adjacency relations:
a r-;:., fi i,j; b
1"/.1
e,h,j; c rp d,g 1 j; d r-;:., h1 i; e r-;:., g,i; and f r-;:., g, h. To detennine
R2 (N5 ) = R(R(N5), we look at C(R.(N5 )) instead of using Definition 2 since R.(N5)
contains a lesser number of edges than R(Ns) and is thus easier to handle. R(N5 )
is the following graph:
c
a
h-:----l.---�
g
The induced chord\ess cycles are A = aidcja, B = aidcgfa , C = afgcja, D =
= afhbja, G = a.idhfa, H = aiegfa, I = die gcd, J =
diebjd, !( = hdcgfh, L = cjbeidc, M = bjcgfhb, N = bhfgeb and O = idhfgei.
aiebja., E = aiebja, F
Then in C(R(N5 )) = R.2(N,1 ), A "' N, C "' J, E "' !( and F "' 1 so that R'(N5 ) is
N
A/
c ...--' J
K
E /
•
B
.,/°
F
I
ol
• • 0•
D G •
H
•
M
90
and R3{Ns) = R(R2 (N,)) = C(R2 (Ns)) = 0.
Therefore, we have shown that Nn is R-vanishing for n < 6.
We now prove that Nn is R-expanding for n
2: 6 as follows;
We apply Proposition
43 to the relation given in Corollary 64: R2m(Nn) r> Nn+2m , m ;:=: 1, n
get R2m+1 (Nn)
I>
2:
6 to
R(Nn+2m ) so that by Proposition48, #1, we have R2m+1 (Nn )
I>
G(n + 2m,3,2), m 2 1, n 2 6. This means that as m --> oo, IV(R2m)(N.) I,
IV(R2m+l )(Nn ) I --> oo. Thus, Nn is R-expanding for n 2 6 by definition. •
The next proposition gives the minimum order of a path for which it is R-expanding.
Proposition 66 The path ?ii of order n is R-vanishing for n < 7 and R-expanding
for n 2 7.
Proof. To prove that Pn is R-vanishing for n < 7, it suffices to show by Proposition
62, #1 that P6 is R-vanishing since P6 t> Pn for n < 7. Let P6
=
123456, then the
vertices of R(P6 ) are a = {l, 3, 5}, b = { 1, 3, 6), c = {2, 4, 6), r1 = {l, 4, 6},
e = {12, 45), f = { 12, 5 6), and g = {23, 56} with the following non-adjacency
relation: a ..;., c1 d; b r,u c1 e; d ,..,,., g so lhat R{Pu) is the graph below:
e
a
b
c
d
g
Thus, R2 (P6) = C(R(P,)) = 0 so that P, is R -vanishing.
Therefore, we have shown that Pn is R-vanishing for n � 7.
To prove that Pn is R-expanding for n � 7, it suffices to show by Proposition 62,
#2 that P7 is R -expanding since Pn I> P1 for n � 7. We let P1 = 1234567, then
a = {1, 3, 5}, b = {l, 3, 6}, c = {1, 3, 7}, d = {l, 4, 6}, e = {l, 4, 7), f = {1,
5, 7}, g = {2, 4, 6}, h = {2, 4, 7}, and i = {2, 5, 7) are 3-pointed stars in P1
91
and m
=
{23 , 56} is a 4-pointed star in P7 and so they are vertices in R(P7). We
observe that in R(P1), a
,y.,
d, e , g , h; b "'"' e,f,g,i; c
r/J
d, g, i, m; d
f';f.l
h,m;
e "" 9,i; f "" g; g "" i so that {a, d, h}, {a, e, g}, {b, e, i}, {b, f, g}, {c, d, m},
and {c, g, i) are six 3-pointed stars in R(P7) any two of which do not have two
vertices in common. Then these six 3-pointed stars form six pairwise non-adjacent
vertices in R2 (P7) so that R 2(P,) t> N6 • This implies that R2 (P7) is R-expanding by
Proposition 62, #2, then so is P1 . Therefore, we have shown that Pn is R-expanding
for n > 7. •
The proposition below tells us that the minimum order of a cycle for which il is
R-expanding is the same as that of a path.
Proposition 67 The chordless cycle
Cn
of order n is R-vani:diing for n
<7
and
R-expanding for n > 7.
Proof. We first show that Cn is R-vanishing for n < 7.
For n = 1, 2,
Cn
is undefined.
It is easy to see that R(c3 ) = R(c,) = 0 and R(c5 ) =
For n = 6, we use the fact that R(c6 )
=
[{1 so that R2 (c5) = 0.
C(s6 ). We have seen in Theorem 16 that
the only non-adjacent vertices in C(s6) are A and C. Thus, R2 (c6 ) = 0.
For n ?.: 8, Cri t> Pn-t and so it follows with Proposition 62 and Proposition 66 that
c,. is R-expanding for n � 8.
It remains lo show that c7 is R-expanding. Let c7 = 12345671, then a = {!, 3, 5},
b = {l, 3, 6} , c = {l, 4, 6}, d = {2, 4, 6), e = {2, 4, 7), f = {2, 5, 7), g
= {3, 5,
7}, h = { 12, 45}, m = {23, 67}, n = {34, 67} and p = {45, 71 }are vertices in R(c,)
with the following non-adjacency relation: a "" c, d, e, f, m, n; b r]{J d, e1 J, g, h, p;
c ""-' e, f, m; d ""-' f, g,p; e "" g; f r]{J n; g � h. Hence, the elements of each of the
six sets A, B, C, D, E and F below are 3-pointed stars in R(c7) and so are vertices
92
in R2( c1) :
A
-
B
-
c
-
D
-
E
F
{ {a,c,e}, {a,d, J }, (b, d,p) }
( {a,c,e}, {a,f,n}, (b, d,g) }
{ {a, c,m}, {a, f,n} , {b,d,f} }
{ {a, c,m}, {a,d,J}, {b, d,g} )
{ {a, c, J } , {b, d,p), {b, e,g) }
( {a,c,J}, {b,d,J}, {b, g , h} }
These six sets correspond to 3-pointed stars in R2( c7 ) . Furthermore, they are vertices
in R3( c7) which form N6 since they are pairwise non-adjacent. Therefore, we have
shown that R3 ( c7) !> N6 • And so by Proposition 62, #2, we have R3 ( c7) is R­
expanding and so is c7. •
Finally, we classify a complete bipartite graph as either R-vanishing or R-expanding
based on the cardinalities of its partite sets.
Proposition 68 l.£t ](( m, n) be a complete bipartite graph with partite sets V1 and
Vi such that IVil
=m
and IVil
= 11.1 m 1 n
?: 1, m < n. Then K(m, n) is R-vanishing
all other values of m and n.
for m = 1, 2 and 1 ::; n :5: 5,· m = n = 3; and m = 3, n = 4; and R-expanding for
Proo£ Note that K(m,n) � K(n,m). We may thus consider only the case where
m :::; n and the same result will hold for m > n.
show that 1(( 2, 5) is R-vanishing as K( 2, 5) � K(m, n) for m = 1, 2 and 1 < n :S 5.
1, 2 and 1 � n < 5, it suffices to
To prove that K(m, n) is R-vanishing for m
=
By Theorem 47, R( I(( 2, 5)) = R(N, + N,)
R(N2 ) U R(N5)
=
=
R(N,) and hence
by Proposition 65, K( 2 1 5) is R-vanishing. Thus, the result follows by Proposition
62, #!.
Since R( K( 3 , 4)) = N1 U 1(4, it follows that R2(K(3, 4)) = C(S,) = 0. This implies
93
that K(3,4) is R -vanishing. Also, note that K(3,4) C> K(3 , 3), and so K(3, 3) is
also R-vanishing by Proposition 62, #!.
Now, K(m, n) is clearly R-expanding if eitherm or n is at least six since R(K(m, n)) =
R(Nm) (J R(Nn) which is R-expanding in this case by Proposition 65.
We are left with the following complete bipattite graphs: K(3, 5), K(4, 4), K(4, 5)
and K(5, 5).
We observe that R(K(3 , 5)) = C(K3 (J K5)
= N1 (J R(N5) = N1 + R(N5) . Letting
x denote the vertex corresponding to N1 and using the labelling of R(N5 ) from the
proof of Proposition 65, it follows that xdhx, xeix, xafx, xbmx, hbegfh and aidcma
that in R2 (!{(3 , 5))
C(R(K(3 , 5) ), they form six pairwise non-adjacent vertices.
are chordless cycles in R{J((3,5)) no two of which have an edge in common so
Therefore, R 2 (!{(3, 5)) C> N0 . Then it follows that R 2(I<(3, 5)) is R-expanding by
=
Proposition 65 and Proposition 62, #2 so that I<(3, 5) is R-expanding, too.
Also, J<(4,5), I<(5,5) C> K(3,5) so that I<(4,5) and I<(5,5) are R-expanding.
Let the partite sets of K(4,4) be
Corollaty 56, R2 (K(4,4))
=
Vi =
{1, 2, 3,4) and V, - {5, 6,7, 8). Then by
C(/{(4,4)). We then look at some of the chordless
cycles of K(4,4), namely, a = 15261, b = 15271, c = 15281, d = 37483, e
= 35463,
tices in R2 (K(4,4)} with lhe following non-adjacency relations: a "" d, e, f, g,h,i;
f = 35473, g
= 35483, h = 36473 and i = 36483. These cycles correspond to ver­
b "" d, e,f, g , h, i; c "" d, e, f,g J i,i; d "" e; J '"P i; and g "" h. Henct,, the ele­
ments of the six sets below are 3-pointed stars in R2 (K(4, 4)) and so are vertices in
R3(K(4,4)) :
A
-
B
c
D
-
{ {a, f ,i}, {a, g , h}, {a,d,e) }
{ {b,J,i ), {b, g, h} , {b, d, e) )
{ {c,f,i}, {c, g , h}, {c,d,e} )
{ {a, J, i}, { b, g , h}, {c, d,e} }
94
E
{ {b, f,i}, {c,g,h}, {a, d,e} }
F
{ {c,f,i}, {a,g,h}, {b, d,e} }
While each of A, B and C intersects each of D, E and F in exactly one element, A, B
and C are pairwise disjoint; and D, E and F are pairwise disjoint. Thus, they form
six pairwise non-adjacent vertices in R'(K(4,4)). Therefore, R'(K(4, 4) ) I> N6 •
Then it follows that K(4, 4) is R-expanding. •
As an immediate consequence of Proposition 68, we have the following corollary
which states tbat the minimum order of a star for which it is R-expanding is six.
Corollary 69 The star S,. cif order n is R-vanishing for n < 6 and R-expandingfor
n 2'. 6.
Proo[ By definition, Sn = K(l, n). And by Proposition 68, K(l, n) is R-vanishing
for n < 6 and R -expanding for n 2 G. Then the result fo!lows. •
95
Chapter 5
Summary, Discussion and
Enumeration of Open Problems
For any given graph G, the cycle graph C(G) of G is the graph whose vertices are
the chordless cycles of G and two vertices in C( G) are adjacent if and only if the
corresponding chordless cycles have at lea'it one edge in common.
The main purpose of this study is to find a relationship between a graph G and the
cycle graph C(G) of its complement.
This dissertation starts with an introductory chapter which recalls some basic notions
and terms in graph theory used in this study and presents some important concepts,
theorems and corollaries concerning cycle graphs and planarity referred to in the
succeeding chapters. Chapter I also discusses the open problem in detail and gives
a summary of the steps undertaken in the next chapter to come up with the desired
solution.
Chapter 2 contains the main result of this study. It introduces the concepts of the
n-pointed star and the regular graph of a graph. The n-pointed star is a graph
possessing certain properties which are later shown to be the properties characterising
the complement of a chordless n-cycle, n 2: 3. On the other hand, the regular
graph of a graph G is the intersection graph whose vertex set is the union of an
induced k-pointed star subgraph sk of G and all pairs of non-adjacent vertices of
sk, k = 3) 41 ••• 1 IV(G)\. The main result contained in this chapter suggests that the
cycle graph of the complement of a given graph can be determined by getting its
regular graph, C(G) 9' R(G) for any graph G, which is the required solution to the
96
problem.
Chapter 3 explores some interesting attributes of then -pointed star with emphasis on
those leading to results concerning cycle graphs which are useful in the succeeding
chapter. It examines some properties of the cycle graph G( sn ) of then-pointed star
for n � 6 which includes the determination of the degree of an arbitrary vertex in
C( sn ), as well as the order IV( C{sn ))I of C( sn )·
The main result of Chapter 2 induces the mapping R : G - C( G) which is defined
for all graphs G. The last chapter, Chapter 4, investigates this mapping and comes up
with results concerning the basic properties of R; the images under R of some special
graphs such as Nn , I<n , K( m1 n), Sn, Hln, Cn , Pn and graphs obtained by performing
some finite graph operations on the n -pointed star and the complete graph. It also
looks at classifying a graph G as either an R-vanishing or an R-persistent graph
based on the iterated regular graph of G. A special type of R-persistent graph is the
R-expanding graph in which lim IV( R"( G)I � oo. Section 4.3 deals with finding
k->oo
the minimum order of a null graph, a path, a cycle and a complete bipartite graph
for which they are R-expanding.
Two contributions of this dissertation to graph theory are the fo!lowing characterisa­
tion of the complement of a chordless cycle and characterisation of the complement
of a forest:
A graph G is the complement of a chordlessn-cycle,n � 3 if and only if G satisfies
the following conditions:
( 1 ) G is (n - 3)-regular of order n.
( 2 ) For n 2: 6, G does not contain K(m,n - m} for any m, 3 :S m
:=:::;
n - 3 as a
subgraph.
A grnph G is the complement of a forest if and only if G does not contain any
n-pointed star, n � 3.
97
Also, having introduced two new graphs, namely; then-pointed star and the regular
graph of a graph, this dissertation opens the door to a number of topics and problems
in graph theory.
The regular graph of a graph is a kind of intersection graph. As often done with
intersection graphs, the inverse operation of associating with a graph G a graph H
with the property that its regular graph is G, ie R( H) "' G, may also be considered.
An obvious related open problem which can then be investigated is: For what graph
G does H exist, that is, find a characterisation of a graph cJ which is the regular
graph of some graph. It might also be worthwhile to look for an application of this
graph as intersection graphs are known to have been used as mathematical models
in representing some structural entities in several fields.
Chapter 3 explores a few properties of the cycle graph C( sn) of an n-pointed star Sn
and fails to obtain a characterisation of the cycle graph C( s11. ) of nn arbitraryn-pointed
star, n � 6. It is likely that further work on this may find such a characterisation.
In this regard, it might be helpful to write a computer program for the algorithm for
finding C( sn),
The last portion of Chapter 4 deals with R-vanishing and R-expanding graphs. These
definitions are analogous to the definitions of cycle-v:mishing and cycle-expanding
graphs. Corresponding to the cycle-periodic graph, an R p- criodic graph may also
be defined as follows: A graph G is R -periodic if there exist a positive integer p
and a non-negative integer q such that RP+q( G) '.::::'. Rq(G). It is shown that some
special graphs such as null graph, path, cycle and complete bipartite graph are either
R-vanishing or R-expanding graphs. Hence, it is natural to ask if this observation
can be generalised, ie are all graphs either R-vanishing or R-expanding? If it is so,
then R-periodic graphs do not exist. Thus, this situation boils down to the problem
of showing the existence or non-existence of an R -periodic graph. It can be observed
that the third and fourth iterated regular graph of the special graphs considered in this
section play an important role in establishing whether such graphs are R-vanishing or
98
R -expanding : If G is some null graph, path, cycle or complete bipartite graph, then
{i)
G is R -vanishing whenever R3 (G)
=
0 and {ii)
G is R-expanding whenever
R'(G) or R4(G) is. It can be recalled that Egawa, Kano and Tan (1991) produced a
characterisation of cycle-vanishing and cycle-expanding graphs in terms of the third
iterated cycle graph of the given graph. Similarly, it may be possible to construct an
analogous characterisation of R-vanishing and R-expanding graphs in tenns of the
third or the fourth iterated regular graph of a given graph.
99
Chapter 6
List of References
Akiyama, J., & Ruiz, M-J. (1985). Intersection graphs and their applications.
Matimyas Matematika 9, 8 3- 3.
Buckingham, M.A. (198 1). Circle graphs. Doctoral dissertation, New York Uni­
versity, USA.
Egawa, Y., Kano, M. & Tan, E.L. (1991). On cycle graphs. Ars Combinatoria
32, 91-113.
Fowler, R., Paterson, M.S. & Tanimoto, S.L. (198 1). Optimal packing and cover­
ing in the plane are NP-complete. brfon11atio11-Processing-l.etters 12 (3) 133 - 1 37.
Gervacio, S.V. (1984). Cycle graphs. In K.M. Koh & H. P. Yap (Eds.). Pro­
ceedings of the First Southeast Asian Graph Theory Colloquium. (pp.279 -293).
Singapore: Springer-Verlag.
Harary, F. (1969). Graph theory. Reading, MA: Addison-Wesley Publishing
Company.
Le, V.B. & Prisner, E. ( 1992). On inverse problems for the cycle graph operator.
Graphs and Combinatorics 8 (2), 155.
Rosenkrantz, D.J. (1995, March). Simple heuristics for unit disk graphs:
Maranathe, M. V., Breu, H., Hunt III, H.B. & Ravi, S.S. (Eds.). Networks 25,
59 -68.
100
Tan, E.L. (1987). On the cycle graph of a graph and inverse cycle graphs. Doc­
toral dissertation, University of the Philippines, Diliman, Quezon City, Philip­
pines.
Tan, E.L. (1993). Some solved and unsolved problems in cycle graphs. In J.
Akiyama, M-J P. Ruiz & T. Saito (Eds.). Proceedings of the JSPS-Workshop,
Tokyo. Japan, 15-/6 August /990 (pp. 8 1-89). Philippines: Mathematics Depart­
ment, Ateneo de Manila University.
Tan, E.L. & Gervacio, S.V. (1993). Necessary conditions for cycle graphs. In H.P.
Yap, et al (Eds.). Proceedings of the Spring School and Intemational Conference
on Combinatorics, Hefei, 6-27 April 1992 PROC (pp. 183-189). Singapore: World
Scientific Publishing Co.